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Due to the nature of estimations based on samples, statistics cannot be guaranteed to be true. On the other hand, how can it be determined whether a certain *claim* about the mean of a population is valid? This is where the two main methods of *inferential statistics* come in action. These are *confidence intervals* and *hypothesis testing*. ### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**
Background to help understand Probability

Challenge

Mark's father runs a burger restaurant. The mean age of people who visit the restaurant is $24.3$ years old. Mark suspects that this situation has changed during the last year. To investigate whether his suspicions were true, he surveyed $65$ customers and found a sample mean of $25.5$ years with a standard deviation of $5$ years.

If he wants to test his results with a $10%$ significance, help him complete the following questions.

a Select the tests of significance needed to make a hypothesis test.

{"type":"choice","form":{"alts":["Right-tailed test","Left-tailed test","Two-tailed test"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":2}

b Consider the following graphs.

{"type":"choice","form":{"alts":["<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.7109375em;vertical-align:0em;\"><\/span><span class=\"mord text\"><span class=\"mord Roboto-Bold textbf\">A<\/span><\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.7109375em;vertical-align:0em;\"><\/span><span class=\"mord text\"><span class=\"mord Roboto-Bold textbf\">B<\/span><\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.73046875em;vertical-align:-0.009765625em;\"><\/span><span class=\"mord text\"><span class=\"mord Roboto-Bold textbf\">C<\/span><\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.7109375em;vertical-align:0em;\"><\/span><span class=\"mord text\"><span class=\"mord Roboto-Bold textbf\">D<\/span><\/span><\/span><\/span><\/span>"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}

c What is the $z-$value of the sample mean? Round to two decimal places.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.48312em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.04398em;\">z<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2248<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["1.93"]}}

d Which statement is more likely true about the mean age of the population of people who eat at the burger restaurant?

{"type":"choice","form":{"alts":["The mean age of the population is not <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">2<\/span><span class=\"mord\">4<\/span><span class=\"mord\">.<\/span><span class=\"mord\">3<\/span><span class=\"mord\">.<\/span><\/span><\/span><\/span>","The mean age of the population is <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">2<\/span><span class=\"mord\">4<\/span><span class=\"mord\">.<\/span><span class=\"mord\">3<\/span><span class=\"mord\">.<\/span><\/span><\/span><\/span>"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}

Discussion

Inferential statistics uses data from a sample to draw conclusions or test hypotheses about a population. Conclusions made from a sample are almost never $100%$ accurate but can be thought of as the best guess or most probable answer. One of the main tasks of inferential statistics is to provide a *confidence interval*.

Concept

The maximum error of estimate, also known as the margin of error, is the maximum difference between the estimate of the population mean $xˉ$ and its actual value. The maximum error of estimate $E$ is calculated using the following formula.

$E=z⋅n s ,n≥30$

In this formula, $z$ represents the $z-$value of a certain confidence level, $s$ is the standard deviation of the sample, and $n$ is the sample size. From the formula, some conclusions can be made about the error of estimate.

- Increasing the sample size while the standard deviation remains the same will result in a smaller margin of error.
- Conversely, an increase in the standard deviation while the sample size remains the same will cause a bigger margin of error.
- The greater the absolute value of $z$ — meaning an increase in the confidence level — the greater the margin of error.

The maximum error of estimate is added to and subtracted from the estimation mean $xˉ$ to find the bounds of a confidence interval.

Concept

A statistic is rarely equal to the population parameter. Due to this uncertainty, estimations are commonly presented as a confidence interval. This is a range of values that the actual parameter is expected to fall within with some degree of certainty. A confidence interval is found by adding and subtracting the maximum error of estimate $E$ to and from the statistic, like the sample mean $xˉ.$
### Confidence Level and the Standard Normal Distribution

The confidence level matches the percentage of the area under the standard normal curve around the mean limited by the $z$ and $-z$ values, as shown below. ### Confidence Interval for the Population Mean

It is worth noting that increasing the level of confidence results in a wider interval that is more likely to catch the true mean, but it will be less precise because it will cover a greater range of values. This means there is a trade-off between confidence and precision.

The degree of certainty, or the confidence level, is usually presented as a percent value. It refers to the reliability of the analysis to produce accurate intervals. For example, if $10$ confidence intervals are produced using different samples of the same size with $90%$ confidence, then $9$ out of $10$ intervals are expected to contain the actual mean.

For a $99%$ confidence interval, there is a $1%$ probability of observing a value outside this area. Because the distribution is symmetric, half of this area will be on each tail of the distribution.

A confidence interval for the population mean can be found by adding and subtracting the maximum error of estimate $E$ to and from the sample mean $xˉ.$

$CI=xˉ±E$

Example

Mark's father owns a burger restaurant. He wants to implement changes to improve the customer experience. Recently he found that in a sample of $36$ burgers, on average, a burger takes $22$ minutes to be cooked and given to the customer, with a standard deviation of $6.2$ minutes.

Use this information to calculate the maximum error of estimate with a $90%$ confidence level. Round the answer to two decimal places.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.05764em;\">E<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2248<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["1.70"]}}

Is the sample size greater than $30?$

Consider the formula for the maximum error of estimate.
### Finding the $z-$value

### Evaluating the Formula

It is worth noting that the formula for the maximum error can be used because the sample size $n=36$ is greater than $30.$ Recall the standard deviation $s=6.2$ and $z=1.645.$ Substitute these values into the formula to find the maximum error of estimate.
Therefore, the maximum error of estimate at a $90%$ confidence level is about $1.70.$

$E=z⋅n s ,n≥30 $

In this formula, $z$ corresponds to the $z-$value of a particular confidence level, $s$ is the standard deviation of the sample, and $n$ is the sample size. To find the maximum error of estimate for this situation, the $z-$value will be determined first. Then the formula will be evaluated. Since the confidence level $c$ is $90%,$ this portion of the area around the mean $μ$ will be covered in a standard normal distribution. The area in the distribution's tails that are not in the confidence interval will be $(100−90)/2=5%$ each.

Because the distribution is symmetric, the $z-$values limiting this area are opposites, so only one value needs to be found. Additionally, this value is given by the $z-$value of the upper or lower tail. One way to determine this value is to use a graphing calculator. Push $2nd ,$ then $VARS ,$ and choose the third option, invNorm(.

Next, enter $0.05$ and push $ENTER $ to get the $z-$value of the lower tail.

The $z-$value is approximately $-1.645,$ and because of the symmetry of the distribution, this means that its additive inverse $1.645$ can be used to evaluate the formula.

$E=z⋅n s $

SubstituteValues

Substitute values

$E=1.645⋅36 6.2 $

MoveLeftFacToNum

$a⋅cb =ca⋅b $

$E=36 1.645⋅6.2 $

Multiply

Multiply

$E=36 10.199 $

CalcRoot

Calculate root

$E=610.199 $

UseCalc

Use a calculator

$E=1.699833…$

RoundDec

Round to $2$ decimal place(s)

$E≈1.70$

Example

The secret to the success of the burger restaurant is not only the flavor of the meat but also the soda included in the King's Combo. This soda follows a unique brewing process, and a soda dispensing machine fills the bottles that are later sold with the combos.

Mark wants to find the mean volume contained in the bottles that are filled by the dispensing machine. He took a sample of $50$ bottles of soda and measured their volumes. He found that the mean volume of the bottles is $330$ milliliters with a standard deviation of $10.$ Which option corresponds to a $99%$ confidence interval for the population mean $μ$ of soda volume?{"type":"choice","form":{"alts":["<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.78041em;vertical-align:-0.13597em;\"><\/span><span class=\"mord\">3<\/span><span class=\"mord\">2<\/span><span class=\"mord\">6<\/span><span class=\"mord\">.<\/span><span class=\"mord\">4<\/span><span class=\"mord\">0<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2264<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.8304100000000001em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\">\u03bc<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2264<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">3<\/span><span class=\"mord\">3<\/span><span class=\"mord\">3<\/span><span class=\"mord\">.<\/span><span class=\"mord\">6<\/span><span class=\"mord\">0<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.78041em;vertical-align:-0.13597em;\"><\/span><span class=\"mord\">3<\/span><span class=\"mord\">2<\/span><span class=\"mord\">7<\/span><span class=\"mord\">.<\/span><span class=\"mord\">2<\/span><span class=\"mord\">3<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2264<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.8304100000000001em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\">\u03bc<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2264<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">3<\/span><span class=\"mord\">3<\/span><span class=\"mord\">2<\/span><span class=\"mord\">.<\/span><span class=\"mord\">7<\/span><span class=\"mord\">7<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.78041em;vertical-align:-0.13597em;\"><\/span><span class=\"mord\">3<\/span><span class=\"mord\">2<\/span><span class=\"mord\">7<\/span><span class=\"mord\">.<\/span><span class=\"mord\">6<\/span><span class=\"mord\">7<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2264<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.8304100000000001em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\">\u03bc<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2264<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">3<\/span><span class=\"mord\">3<\/span><span class=\"mord\">2<\/span><span class=\"mord\">.<\/span><span class=\"mord\">3<\/span><span class=\"mord\">3<\/span><\/span><\/span><\/span>"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}

Begin by calculating the maximum error of estimate. Then add and subtract that from the sample mean to get the bounds of the confidence interval.

Determine a confidence interval for the population mean $μ$ of soda volume in order to identify the right option. To do so, follow these steps.

- Identify the sample mean.
- Calculate the maximum error of estimate.

The mean volume for the sample consisting of $50$ sodas was $330$ milliliters. The maximum error of estimate will be calculated next.

$E=z⋅n s ,n≥30 $

In this formula, $z$ corresponds to the $z-$value of the confidence level. Since the confidence level is $99%,$ this portion of the area around the mean $μ$ will be covered in a standard normal distribution. The area in the distribution's tails that are not in the confidence interval will be $(100−99)/2=0.5%$ each.
This value is given by the $z-$value of the upper or lower tail. Because the distribution is symmetric, the $z-$values limiting this area will be opposites of each other, so only one needs to be found. In this case, a short version of the standard normal table can be used to locate the $z-$value of the lower tail, which in decimal form is $0.005.$

$.0$ | $.1$ | $.2$ | $.3$ | $.4$ | $.5$ | $.6$ | $.7$ | $.8$ | $.9$ | |
---|---|---|---|---|---|---|---|---|---|---|

$-3$ | $.00135$ | $.00097$ | $.00069$ | $.00048$ | $.00034$ | $.00023$ | $.00016$ | $.00011$ | $.00007$ | $.00005$ |

$-2$ | $.02275$ | $.01786$ | $.01390$ | $.01072$ | $.00820$ | $.00621$ | $.00466$ | $.00347$ | $.00256$ | $.00187$ |

$-1$ | $.15866$ | $.13567$ | $.11507$ | $.09680$ | $.08076$ | $.06681$ | $.05480$ | $.04457$ | $.03593$ | $.02872$ |

$-0$ | $.50000$ | $.46017$ | $.42074$ | $.38209$ | $.34458$ | $.30854$ | $.27425$ | $.24196$ | $.21186$ | $.18406$ |

$0$ | $.50000$ | $.53983$ | $.57926$ | $.61791$ | $.65542$ | $.69146$ | $.72575$ | $.75804$ | $.78814$ | $.81594$ |

$1$ | $.84134$ | $.86433$ | $.88493$ | $.90320$ | $.91924$ | $.93319$ | $.94520$ | $.95543$ | $.96407$ | $.97128$ |

$2$ | $.97725$ | $.98214$ | $.98610$ | $.98928$ | $.99180$ | $.99379$ | $.99534$ | $.99653$ | $.99744$ | $.99813$ |

$3$ | $.99865$ | $.99903$ | $.99931$ | $.99952$ | $.99966$ | $.99977$ | $.99984$ | $.99989$ | $.99993$ | $.99995$ |

$20.00621+0.00466 ≈0.005 $

This means that the $z-$value of $0.005$ can be approximated by finding the mean of the two $z-$values, $-2.5$ and $-2.6.$
$2-2.5+(-2.6) ≈-2.55 $

Since the $z-$values of the distribution are additive inverses, the positive value can be used to evaluate the formula and determine the maximum error of estimate. The formula can be used because the sample size $n=50$ is greater than $30.$ Recall that the standard deviation is $s=10.$
$E=z⋅n s $

SubstituteValues

Substitute values

$E=2.55⋅50 10 $

▼

Simplify right-hand side

MoveLeftFacToNum

$a⋅cb =ca⋅b $

$E=50 2.55⋅10 $

Multiply

Multiply

$E=50 25.5 $

UseCalc

Use a calculator

$E=3.606244…$

RoundDec

Round to $2$ decimal place(s)

$E≈3.60$

$CI=xˉ±E $

Consider the positive and negative cases to determine the bounds of the confidence interval. $CI=xˉ±E$ | |
---|---|

$xˉ−E$ | $xˉ+E$ |

$330−3.60$ | $330+3.60$ |

$326.40$ | $333.60$ |

$326.40≤μ≤333.60 $

Discussion

While a confidence interval helps estimate the value of a population parameter like the mean, there is another inferential method that can help evaluate a specific claim about a population parameter. Before exploring this method, two statistical hypotheses about the population need to be identified. These are the *null* and *alternative hypotheses*.

Concept

The null hypothesis and alternative hypothesis are two mutually exclusive statements about the mean of a population. The null hypothesis, denoted by $H_{0},$ is a statement of equality or non-strict inequality about the population mean that is accepted as true unless strong evidence is shown against it.

$H_{0}:$ Null Hypothesis

Conversely, the alternative hypothesis, denoted by $H_{a}$ or $H_{1},$ is a strict inequality statement that contradicts the null hypothesis. It is the *complement* of the null hypothesis and will be accepted if there is evidence in its favor.

$H_{a}:$ Alternative Hypothesis

Notice that the initial claim made by the researcher is the one that sets the null and alternative hypotheses. If the claim can be written algebraically as a strict inequality, it will be part of the alternative hypothesis. Otherwise, it will be part of the null hypothesis.

$μ>3.2 $

Because this is a strict inequality, the claim represents the alternative hypothesis, while the null hypothesis is $μ≤3.2.$ - $H_{0}:μ≤3.2$
- $H_{a}:μ>3.2$

Example

Another characteristic of the King's Combo at Mark's father's restaurant is that customers can choose between a cookie or a soft ice cream as part of their meal. They can also pay $$2$ more to get a piece of cake.

a Mark thinks that on a typical day, less than $60%$ of customers choose the cookie over the ice cream. Which of the following options describe the alternative hypothesis $H_{a}$ and the null hypothesis $H_{0}$ of this situation?

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class=\"mrel\">:<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.625em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\">\u03bc<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">0<\/span><span class=\"mord\">.<\/span><span class=\"mord\">6<\/span><span class=\"mord\">0<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.83333em;vertical-align:-0.15em;\"><\/span><span class=\"mord\"><span class=\"mord mathdefault\" style=\"margin-right:0.08125em;\">H<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.151392em;\"><span style=\"top:-2.5500000000000003em;margin-left:-0.08125em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">a<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">:<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.7335400000000001em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\">\u03bc<\/span><span class=\"mspace\" 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mtight\"><span class=\"mord mtight\"><span class=\"mord mathdefault mtight\">a<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">:<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.8304100000000001em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\">\u03bc<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2264<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">0<\/span><span class=\"mord\">.<\/span><span 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b Suppose that Mark thinks that on average, $50%$ of the customers on a typical day added a piece of cake to their combo. Which of the following options describe the null and the alternative hypothesis?

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style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">:<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.7335400000000001em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\">\u03bc<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">0<\/span><span class=\"mord\">.<\/span><span class=\"mord\">5<\/span><span class=\"mord\">0<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.83333em;vertical-align:-0.15em;\"><\/span><span class=\"mord\"><span class=\"mord mathdefault\" style=\"margin-right:0.08125em;\">H<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.30110799999999993em;\"><span style=\"top:-2.5500000000000003em;margin-left:-0.08125em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mtight\">0<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">:<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.8304100000000001em;vertical-align:-0.19444em;\"><\/span><span class=\"mord 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a Begin by identifying the claim. Then write the claim as an algebraic expression.

b The null hypothesis is a statement of equality or a non-strict inequality. The alternative hypothesis is a statement of strict inequality or compound inequality that is the complement of the null hypothesis.

a Begin by identifying the claim and writing it as an algebraic expression. In this case, Mark suspects that the proportion of people that prefer the cookie over the ice cream is less than $60%.$ This can be represented as the following strict inequality.

$μ<0.60 $

The claim is that the mean is less than $0.60.$ Because it is an strict inequality, it can be an alternative hypothesis. Conversely, the complement of this claim is that the mean $μ$ is greater than or equal to $0.60.$ This is the statement of equality that represents the null hypothesis. Null Hypothesis | Alternative Hypothesis |
---|---|

The mean is greater than or equal to $0.60.$ $H_{0}:μ≥0.60$ |
The mean is less than $0.60.$ (claim) $H_{a}:μ<0.60$ |

b The null and alternative hypotheses can be identified in this situation by following a similar procedure. In this case, Mark thinks that $50%$ of customers add a piece of cake to their combo. This claim can be represented by the following equality.

$μ=0.50 $

Since this is a statement of equality, it is the null hypothesis. On the other hand, the alternative hypothesis is $μ =0.50,$ which is the complement of $μ=0.50.$ Null Hypothesis | Alternative Hypothesis |
---|---|

The mean is equal to $0.50.$ (claim) $H_{0}:μ=0.50$ |
The mean is not equal to $0.50.$ $H_{a}:μ =0.50$ |

Discussion

Once the null and alternative hypotheses have been correctly identified, they can be tested by performing a hypothesis test to see which statement is more likely true. Before the test can be performed, some information is needed.

Concept

A hypothesis test is an inferential method that uses sample data to examine a claim about the mean $μ$ of a population. Because the population mean is almost always unknown, it is common to be suspicious about the truthfulness of any assumption about its value. The following are typical claims about the mean of a population.

Typical Claims About the Mean | ||
---|---|---|

The mean is equal to a specific value, $μ=k.$ | The mean is greater than a specific value, $μ>k.$ | The mean is less than a specific value, $μ<k.$ |

Before making a hypothesis test, two hypotheses need to be specified, the null hypothesis and the alternative hypothesis. These hypotheses must be mutually exclusive. The null hypothesis $H_{0}$ is assumed to be true. The hypothesis test puts the null hypothesis on trial to see if there is strong evidence against it. If so, the alternative hypothesis $H_{a}$ is accepted instead.

- $H_{0}:$ The hypothesis that is examined.
- $H_{a}:$ The hypothesis that is accepted if there is strong evidence to reject $H_{0}.$

Concept

The significance level $α$ is the probability that the results obtained in a sample are due to chance and is set in advance when making a hypothesis test. The smaller the $α$ value, the stronger the results of a sample are. These are typical values for the significance level.

Typical Significance Levels $α$ | ||
---|---|---|

$1%$ | $5%$ | $10%$ |

In a standard normal distribution, the sample mean would fall around the center of the distribution if the null hypothesis $H_{0}$ were true. This means that a value in the tails of the distribution would be unusual if $H_{0}$ were true. The significance level tells how far the sample mean will lie in from the center of the distribution and whether to reject the null hypothesis and accept the alternative hypothesis $H_{a}.$

Concept

The critical region, determined by the significance level $α,$ is the set of values that will lead to rejecting the null hypothesis $H_{0}.$ In a standard normal distribution, this region is located in the tails of the distribution. The cutoff value of the region is a critical value given by the $z-$value of $α.$ The tests of significance — left, right, or two-tail — determine whether there are one or two critical regions.

Critical Values | |||
---|---|---|---|

Significance Level | Left-Tail Test $H_{a}:μ<k$ |
Two-Tail Test $H_{a}:μ =k$ |
Right-Tail Test $H_{a}:μ>k$ |

$α=1%$ | $-2.326$ | $±2.576$ | $2.326$ |

$α=5%$ | $-1.645$ | $±1.960$ | $1.645$ |

$α=10%$ | $-1.282$ | $±1.645$ | $1.282$ |

Concept

In a hypothesis test, the region where the null hypothesis is rejected is known as the critical region. The location of this region depends on the significance level $α$ and the inequality symbol of the alternative hypothesis as determined by the tests of significance. The tests of significance can be divided into the left-tailed test, the two-tailed test, and the right-tailed test.

**Left-tailed test**: The alternative hypothesis $H_{a}$ suggests that the population mean is*less than*the value $k$ claimed by the null hypothesis, $H_{a}:μ<k.$**Two-tailed test**: The alternative hypothesis claims that the population mean is different from the value claimed by the null hypothesis, $H_{a}:μ =k.$**Right-tailed test**: The alternative hypothesis suggests that the population mean is*greater than*the value $k$ claimed by the null hypothesis $H_{a}:μ>k.$

The applet below shows how the critical regions vary depending on the tests of significance.

Method

When making a hypothesis test, begin by identifying the claim to set the null and alternative hypotheses. Then the critical regions and the critical values are determined based on the tests of significance. Finally, the null hypothesis is rejected if the $z-$statistic falls within the critical region. To illustrate this process, consider the following situation.

A company says that each of their packages of ham contains exactly $20$ slices. |

1

Identify the Claim and State the Null and Alternative Hypotheses

Identify the claim to see if it relates to the null or the alternative hypotheses. In this case, the company says that the packages contain exactly $20$ slices of ham. This is the same as stating that the population mean $μ$ equals $20,$ which can be written as follows.

$μ=20 $

Because this claim is a statement of equality, it should be related to the null hypothesis $H_{0}.$ Conversely, the alternative hypothesis $H_{a}$ is $μ =20,$ which is the complement of $μ=20.$ Null Hypothesis $H_{0}$ | Alternative Hypothesis $H_{a}$ |
---|---|

The mean is equal to $20$ slices (claim). $H_{0}:μ=20$ |
The mean is different than $20$ slices. $H_{a}:μ =20$ |

2

Determine the Critical Value(s) and Critical Region(s)

Because the sign of the alternative hypothesis is $ =,$ a two-tailed test of significance will be conducted. This means that there are two critical regions whose cutoffs will be given by the $z-$value of the significance level $α.$ The following are the critical values for the most common $α$ values.

Critical Values | |||
---|---|---|---|

Significance Level | Left-Tail Test $H_{a}:μ<k$ |
Right-Tail Test $H_{a}:μ>k$ |
Two-Tail Test $H_{a}:μ =k$ |

$α=1%$ | $-2.326$ | $2.326$ | $±2.576$ |

$α=5%$ | $-1.645$ | $1.645$ | $±1.960$ |

$α=10%$ | $-1.282$ | $1.282$ | $±1.645$ |

From the table, note that the critical values for a $10%$ significance level are $±1.645.$ Now the critical regions and critical values can be labeled.

3

Calculate the $z-$Statistic

The $z-$statistic — the $z-$value of the sample mean — can be calculated using the following formula.

$z=n s xˉ−μ $

In this formula, $xˉ$ is the sample mean, $μ$ is the population mean, $s$ is the standard deviation of the sample, and $n$ is the sample size. For the given example, it is given that $xˉ=19.5,$ $μ=20,$ $s=2,$ and $n=100.$
$z=n s xˉ−μ $

SubstituteValues

Substitute values

$z=100 2 19.5−20 $

▼

Evaluate right-hand side

SubTerm

Subtract term

$z=100 2 -0.5 $

DivByFracD

$b/ca =ba⋅c $

$z=2-0.5⋅100 $

CalcRoot

Calculate root

$z=2-0.5⋅10 $

MoveNegNumToFrac

Put minus sign in front of fraction

$x=-20.5⋅10 $

Multiply

Multiply

$x=-25 $

CalcQuot

Calculate quotient

$z=-2.5$

4

Reject or Fail to Reject the Null Hypothesis

Next, verify if the $z-$statistic falls within the critical region. If so, reject the null hypothesis. To do so, plot the $z-$statistic jointly with the critical regions to see where it falls, outside or inside the critical region.

Because the $z-$statistic falls within the critical region, the null hypothesis $H_{0}$ is rejected in this case.

5

Make a Conclusion About the Claim

Use the result of the previous step to make a conclusion about the initial claim.

A company says that each of their packages of ham contains exactly $20$ slices. |

In this case, since the initial claim is related to the null hypothesis, it can be said that there is enough evidence to reject the claim that the packages of ham contain exactly $20$ slices.

The following situations need to be considered when calculating the critical values.

- The $z-$values on the lower half of the distribution will be negative.
- Conversely, the $z-$values on the upper half will be positive.
- Because the standard normal distribution is symmetric, the $z-$values of the lower percent $c$ will be opposite of the $z-$value of the upper percent $100−c.$

For the given example, each critical region will cover an area of $5%.$ Therefore, the $z-$value for the left $0.05$ will be found first. To do so, push $2nd ,$ then $VARS ,$ and choose the third option, invNorm(.

Now enter the desired value, which in this case is $0.05.$ Finally, push $ENTER $ to get the result.

The $z-$value for the left tail is about $-1.645,$ so the $z-$value for the right tail will be $1.645.$ A similar process is followed when performing a one-tail test.

Example

While watching the {"type":"choice","form":{"alts":["Right-tailed test","Left-tailed test","Two-tailed test"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":2}
Select the option that represents the critical region for this hypothesis test.
### Hint

### Solution

Dinos and Dragonsmovie with his family, Mark decides to eat a bar of his favorite chocolate as a snack. After eating it, he feels slightly disappointed because the bar seemed a little smaller than the $150$ grams listed on the packages. He decides to investigate if the brand producing the chocolate bars lied about the weight of the chocolate bars.

To determine if what the package shows is true, Mark weighs a sample of $50$ chocolate bars and finds a sample mean of $148.9g$ with a standard deviation of $5.5g.$ He wants to test the affirmation in the packages about the weight of chocolate bars with $5%$ significance. Help him find the following information to draw a conclusion.

a Which test of significance does Mark need to use to make a hypothesis test based on his sample results?

b Investigate the following graphs.

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c What is the $z-$value of the sample mean? Round to three decimal places.

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d After testing his results, which statement is most probably true about the mean weight of the population of chocolate bars?

{"type":"choice","form":{"alts":["The mean weight of chocolate bars is different than <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8524478125em;vertical-align:-0.2080078125em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mord\">5<\/span><span class=\"mord\">0<\/span><span class=\"mord text\"><span class=\"mord\">\u00a0<\/span><span class=\"mord Roboto-Regular\">g<\/span><\/span><span class=\"mord\">.<\/span><\/span><\/span><\/span>","The mean weight of chocolate bars is <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8524478125em;vertical-align:-0.2080078125em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mord\">5<\/span><span class=\"mord\">0<\/span><span class=\"mord text\"><span class=\"mord\">\u00a0<\/span><span class=\"mord Roboto-Regular\">g<\/span><\/span><span class=\"mord\">.<\/span><\/span><\/span><\/span>"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":1}

a Listing the weight on the package is the same as stating that the mean weight of chocolate bars is $150g.$

b The critical value is given by the $z-$value corresponding to half the significance level.

c Use the formula $z=n s xˉ−μ $ to calculate the $z-$statistic.

d Where does the $z-$statistic fall in the distribution?

a Begin by setting the null and alternative hypotheses. Then, based on the inequality sign of the alternative hypothesis, determine the test of significance. Note that the chocolate bar wrappers claim that the weight of each bar is $150g.$ This is the same as saying that the mean weight $μ$ of the population of chocolate bars is $150g.$

$μ=150g $

Because this is a statement of equality, it is related to the null hypothesis $H_{0}.$ Additionally, the complement of this statement is that the mean is different than $150g.$ This statement corresponds to the alternative hypothesis $H_{a}.$ Null Hypothesis $H_{0}$ | Alternative Hypothesis $H_{a}$ |
---|---|

The mean is equal to $150g.$ (claim) $H_{0}:μ=150g$ |
The mean is different than $150g.$ $H_{a}:μ =150g$ |

Because the sign of the alternative hypothesis is $ =,$ a two-tailed test of significance corresponds to this situation.

b In a two-tailed test, there are two critical regions whose critical values are given by the $z-$value of half the significance level $α.$ In a graphing calculator, push $2nd ,$ then $VARS ,$ and choose the third option,

invNorm(.

Next, given that $20.5 =0.025,$ enter $0.025$ and push $ENTER $ to get the result.

This is the critical value corresponding to the critical region on the left of the standard normal distribution. Because the distribution is symmetric, the critical value for the upper tail will be the same but with the opposite sign. With this information, the critical regions can be set in the distribution.

This corresponds to option **A**.

c Calculate the $z-$statistic using the following formula.

$z=n s xˉ−μ $

In this formula, $xˉ$ is the sample mean, $μ$ is the population mean, $s$ is the standard deviation of the sample, and $n$ is the sample size. In this case, $xˉ=148.9,$ $μ=150,$ $s=5.5,$ and $n=50.$
$z=n s xˉ−μ $

SubstituteValues

Substitute values

$z=50 5.5 148.9−150 $

▼

Evaluate right-hand side

SubTerm

Subtract term

$z=50 5.5 -1.1 $

DivByFracD

$b/ca =ba⋅c $

$z=5.5-1.1⋅50 $

UseCalc

Use a calculator

$z=1.414213…$

RoundDec

Round to $3$ decimal place(s)

$z≈-1.414$

d Now, draw the $z-$statistic into the graph of the critical regions to see where it falls. If it lies in the critical region, reject the null hypothesis.

Note that the $z-$statistic falls outside the critical region. Therefore, the null hypothesis cannot be rejected. This means that there is not enough evidence to reject the claim about the weight of the chocolate bars. So, it is most likely true that the mean weight of the chocolate bars is $150g.$

Example

After enjoying the Dinos and Dragons

movie with his family, Mark and his father start watching sports news. The newscaster reports that, on average, teens spend at most $59$ minutes a day playing sports. Mark wants to determine if what the news reported is accurate.

Using a sample of $35$ teens, Mark calculates a mean of $62$ minutes and a standard deviation of $6$ minutes. Help Mark if he wants to test the news report with $5%$ significance.

a Select the test of significance that Mark needs to make a hypothesis test about the results the news reported.

{"type":"choice","form":{"alts":["Right-tailed test","Left-tailed test","Two-tailed test"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}

b Look at the following graphs.

{"type":"choice","form":{"alts":["<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.7109375em;vertical-align:0em;\"><\/span><span class=\"mord text\"><span class=\"mord Roboto-Bold textbf\">A<\/span><\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.7109375em;vertical-align:0em;\"><\/span><span class=\"mord text\"><span class=\"mord Roboto-Bold textbf\">B<\/span><\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.73046875em;vertical-align:-0.009765625em;\"><\/span><span class=\"mord text\"><span class=\"mord Roboto-Bold textbf\">C<\/span><\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.7109375em;vertical-align:0em;\"><\/span><span class=\"mord text\"><span class=\"mord Roboto-Bold textbf\">D<\/span><\/span><\/span><\/span><\/span>"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":3}

c Calculate the $z-$value corresponding to the upper $5%$ of the distribution. Round the answer to three decimal places.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.48312em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.04398em;\">z<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">\u2248<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["2.958"]}}

d After analyzing his sample results, what is more likely true about the mean time spent by teens playing sports?

{"type":"choice","form":{"alts":["The mean time spent by teens playing sports is less than or equal to <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">5<\/span><span class=\"mord\">9<\/span><\/span><\/span><\/span> minutes.","The mean time spent by teens playing sports is greater than <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">5<\/span><span class=\"mord\">9<\/span><\/span><\/span><\/span> minutes."],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":1}

a Identify the claim to set the null and alternative hypothesis. Next, look at the sign of the alternative hypothesis to determine the appropriate test of significance.

b Calculate the $z-$value corresponding to the upper $5%$ of the distribution.

c Use the formula $z=n s xˉ−μ $ to calculate the $z-$statistic.

d If the $z-$statistic falls in the critical region, reject the null hypothesis.

a To identify the test of significance that applies to this situation, identify the claim. Next, set the null and alternative hypotheses and look at the inequality symbol of the alternative hypothesis. The newscaster claims that the mean time $μ$ spent by teens on sports is at most $59$ minutes.

$μ≤59minutes $

This is a non-strict inequality, meaning that it corresponds to the null hypothesis $H_{0}.$ The complement of this statement is that the mean is greater than $59$ minutes, representing the alternative hypothesis $H_{a}.$ Null Hypothesis $H_{0}$ | Alternative Hypothesis $H_{a}$ |
---|---|

The mean is less than or equal to $59$ minutes. (claim) $H_{0}:μ≤59minutes$ |
The mean is greater than $59$ minutes. $H_{a}:μ>59minutes$ |

Because the sign of the alternative hypothesis is $>,$ a right-tailed test of significance applies to this situation.

b With the information found previously, locate the critical region. First, calculate the $z-$value corresponding to the upper $5%$ of the standard normal distribution using a graphic calculator. Push $2nd ,$ then $VARS ,$ and choose the third option,

invNorm(.

Because the upper $5%$ of the distribution is desired, the value to be entered into the calculator is given by $1−0.05=0.95.$ Next, enter this value and push $ENTER $ to get the result

The critical value is about $1.645.$ This value will limit the critical region that will be located in the right tail of the distribution.

Therefore, this corresponds to option **D**.

c Calculate the $z-$statistic using the following formula.

$z=n s xˉ−μ $

In this formula, $xˉ$ is the sample mean, $μ$ is the population mean, $s$ is the standard deviation of the sample, and $n$ is the sample size. In this case, $xˉ=62,$ $μ=59,$ $s=6,$ and $n=35.$
$z=n s xˉ−μ $

SubstituteValues

Substitute values

$z=35 6 62−59 $

▼

Evaluate right-hand side

SubTerm

Subtract term

$z=35 6 3 $

DivByFracD

$b/ca =ba⋅c $

$z=63⋅35 $

UseCalc

Use a calculator

$z=2.958039…$

RoundDec

Round to $3$ decimal place(s)

$z≈2.958$

d To make a conclusion about the initial claim, plot the $z-$statistic to see its position in the distribution. If it falls in the critical region, reject the null hypothesis.

Since the $z-$statistic falls in the critical region, the null hypothesis should be rejected. Additionally, because the initial claim is related to the null hypothesis, it can be said that it is more likely that the mean time spent by teens playing sports is greater than $59$ minutes.

Closure

This lesson reviewed the importance of samples when it comes to estimating population parameters. However, due to the margin of error in estimations, inferential methods are helpful when stating how confident a specific estimation is or testing a particular claim about the population mean.

Inferential Methods | |
---|---|

Confidence Interval | Hypothesis Test |

Estimates a population parameter as a range of values | Tests a claim about the mean of a population |

Now the challenge presented earlier about the average age of people at the burger restaurant can be solved.

The mean age of people who eat at Mark's father's burger restaurant used to be $24.3.$ Mark suspects that this has changed, so he surveyed a sample of $65$ customers. He found a sample mean of $25.5$ years with a standard deviation of