5. Confidence Intervals and Hypothesis Testing
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Chapter 9
5.
Confidence Intervals and Hypothesis Testing
This lesson aims to simplify two critical statistical concepts: confidence intervals and hypothesis testing. Understanding these topics is vital for anyone involved in research, be it in academia, healthcare, or business. Confidence intervals help you estimate a range within which a particular value is likely to fall, making them useful in predicting election outcomes or determining product quality. Hypothesis testing, on the other hand, allows you to make educated guesses about a population based on sample data. This is particularly useful in fields like medicine for drug testing or in marketing to gauge consumer preferences. Together, these statistical tools empower you to make more accurate and reliable conclusions in various professional settings.
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Lesson Settings & Tools
| | 11 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Confidence Intervals and Hypothesis Testing
Slide of 11
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
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Has the Average Age of People Eating at a Restaurant Changed?
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":["A","B","C","D"],"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":[]},"rendered":false},"text":" "},"formTextBefore":"z\u2248","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 24.3.","The mean age of the population is 24.3."],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
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Using Samples to Make Conclusions About a Population
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.
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Maximum Error of Estimate
The maximum error of estimate, also known as the margin of error, is the maximum difference between the estimate of the population mean x and its actual value. The maximum error of estimate E is calculated using the following formula.
E=z*s/sqrt(n), 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 x to find the bounds of a confidence interval.
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Confidence Level
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 x.
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.
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.
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
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 x.
CI=x± E
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Burger Prep Time
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":" "},"formTextBefore":"E\u2248","formTextAfter":null,"answer":{"text":["1.70"]}}
Hint
Is the sample size greater than 30?
Solution
Consider the formula for the maximum error of estimate. E=z*s/sqrt(n), 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.
Finding the z-value
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.
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.E=z*s/sqrt(n)
SubstituteValues
Substitute values
E= 1.645*6.2/sqrt(36)
MoveLeftFacToNum
a*b/c= a* b/c
E=1.645*6.2/sqrt(36)
Multiply
Multiply
E=10.199/sqrt(36)
CalcRoot
Calculate root
E=10.199/6
UseCalc
Use a calculator
E=1.699833...
RoundDec
Round to 2 decimal place(s)
E≈ 1.70
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Amount of Soda Poured by a Dispensing Machine
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.
{"type":"choice","form":{"alts":["326.40\u2264\u03bc\u2264 333.60","327.23\u2264\u03bc\u2264 332.77","327.67\u2264\u03bc\u2264 332.33"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
Hint
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.
Solution
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.
Calculating the Maximum Error of Estimate
For a sample size greater than or equal to 30 with standard deviation s, the maximum error of estimate at a confidence level c can be calculated by the following formula. E=z*s/sqrt(n), 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 |
E=z*s/sqrt(n)
SubstituteValues
Substitute values
E= 2.55*10/sqrt(50)
▼
Simplify right-hand side
MoveLeftFacToNum
a*b/c= a* b/c
E=2.55*10/sqrt(50)
Multiply
Multiply
E=25.5/sqrt(50)
UseCalc
Use a calculator
E=3.606244...
RoundDec
Round to 2 decimal place(s)
E≈ 3.60
Determining the Confidence Interval
The confidence interval for the population mean can now be calculated by adding and subtracting the maximum error of estimate from the sample mean. CI=x± E Consider the positive and negative cases to determine the bounds of the confidence interval.
| CI = x ± E | |
|---|---|
| x - E | x + E |
| 330 - 3.60 | 330 + 3.60 |
| 326.40 | 333.60 |
Therefore, with 99 % level of confidence, it can be said that the population mean μ is between 326.40 and 333.60 326.40≤μ≤333.60
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Testing an Estimation of a Population Parameter
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.
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Null Hypothesis vs. Alternative Hypothesis
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.
Example
Suppose a school administrator at East Junior High School thinks that the mean grade point average μ in 2012 is greater than 3.2. Note that this claim can be written as the following inequality. 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
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Determining the Null and Alternative Hypotheses From a Claim
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?
{"type":"multichoice","form":{"alts":["H_0:\u03bc=0.60","H_a:\u03bc< 0.60","H_a:\u03bc\u2264 0.60","H_0:\u03bc\u2265 0.60"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":[3,1]}
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?
{"type":"multichoice","form":{"alts":["H_0:\u03bc=0.50","H_a:\u03bc≠ 0.50","H_a:\u03bc>0.50","H_0:\u03bc\u2264 0.50"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":[0,1]}
Hint
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.
Solution
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 |
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Collecting the Tools to Make a Hypothesis Test
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.
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Hypothesis Test
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.
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Significance Level
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.
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Critical Region
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 | 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 |
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Tests of Significance
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
- 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.
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Making a Hypothesis Test
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
expand_more 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)
expand_more 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 | 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
expand_more The z-statistic — the z-value of the sample mean — can be calculated using the following formula.
z=x-μ/ssqrt(n)
In this formula, x 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 x= 19.5, μ= 20, s= 2, and n= 100.
z = x-μ/ssqrt(n)
SubstituteValues
Substitute values
z = 19.5 - 20/2sqrt(100)
▼
Evaluate right-hand side
SubTerm
Subtract term
z = -0.5/2sqrt(100)
DivByFracD
a/b/c= a * c/b
z = -0.5 * sqrt(100)/2
CalcRoot
Calculate root
z = -0.5 * 10/2
MoveNegNumToFrac
Put minus sign in front of fraction
x=- 0.5 * 10/2
Multiply
Multiply
x=- 5/2
CalcQuot
Calculate quotient
z =-2.5
4
Reject or Fail to Reject the Null Hypothesis
expand_more 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
expand_more 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.
Extra
Using a Calculator to Find the Critical ValuesThe 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.
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Evaluate the Weight of Chocolate Bars
While watching the 
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.
Select the option that represents the critical region for this hypothesis test.
z=x-μ/ssqrt(n)
In this formula, x 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, x= 148.9, μ= 150, s= 5.5, and n= 50.
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.
a Which test of significance does Mark need to use to make a hypothesis test based on his sample results?
{"type":"choice","form":{"alts":["Right-tailed test","Left-tailed test","Two-tailed test"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":2}
b Investigate the following graphs.
{"type":"choice","form":{"alts":["A","B","C","D"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
c What is the z-value of the sample mean? Round to three decimal places.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":" "},"formTextBefore":"z\u2248","formTextAfter":null,"answer":{"text":["-1.414"]}}
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 150g.","The mean weight of chocolate bars is 150g."],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":1}
Hint
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= x-μ ssqrt(n) to calculate the z-statistic.
d Where does the z-statistic fall in the distribution?
Solution
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 0.52=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 = x-μ/ssqrt(n)
SubstituteValues
Substitute values
z = 148.9 - 150/5.5sqrt(50)
▼
Evaluate right-hand side
SubTerm
Subtract term
z = -1.1/5.5sqrt(50)
DivByFracD
a/b/c= a * c/b
z = -1.1 * sqrt(50)/5.5
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.
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Time Teens Spend Playing Sports
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":["A","B","C","D"],"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.
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d After analyzing his sample results, what is more likely true about the mean time spent by teens playing sports?
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Hint
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= x-μ ssqrt(n) to calculate the z-statistic.
d If the z-statistic falls in the critical region, reject the null hypothesis.
Solution
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≤59 minutes |
The mean is greater than 59 minutes. H_a:μ> 59 minutes |
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 = x-μ/ssqrt(n)
SubstituteValues
Substitute values
z = 62 - 59/6sqrt(35)
▼
Evaluate right-hand side
SubTerm
Subtract term
z = 3/6sqrt(35)
DivByFracD
a/b/c= a * c/b
z = 3 * sqrt(35)/6
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.
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Concluding if the Average Age of People Eating at the Burger Restaurant Has Changed
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 5 years. If he wants to conduct a test with 10 % significance, help him through the hypothesis test.
a Select the tests of significance needed to make a hypothesis test.
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b Consider the following graphs.
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c What is the z-value of the sample mean? Round to two decimal places.
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d Which statement is more likely true about the mean age of the population of customers at the burger restaurant?
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Hint
a Begin by identifying the claim to set the two hypotheses needed for the hypothesis test.
b The critical value is given by the z-value corresponding to half the significance level.
c Use the formula z= x-μ ssqrt(n) to calculate the z-statistic.
d Draw the z-statistic into the distribution to see if it falls within the critical region.
Solution
a The tests of significance are based on the inequality sign of the alternative hypothesis. Therefore, begin by identifying the claim to set the null and alternative hypotheses. In this case, the claim is that the mean age μ of the population attending the restaurant is 24.3 years.
μ=24.3 Since this is a statement of equality, it represents the null hypothesis H_0. Moreover, the complement of this statement is that the mean is different than 24.3, which represents the alternative hypothesis H_a.
| Null Hypothesis H_0 | Alternative Hypothesis H_a |
|---|---|
| The mean is equal to 24.3. (claim) H_0: μ=24.3 |
The mean is different than 24.3 years. H_a:μ≠24.3 |
Because the sign of the alternative hypothesis is ≠, a two-tailed test of significance will be needed in this case.
b In this situation, there will be two critical regions whose critical values are given by the z-value of the significance level α. The critical value of the left tail will be determined by using a graphing calculator. First, push 2nd, then VARS, and choose the third option,
invNorm(.
Mark wants to test his hypothesis at a 10 % significance level, meaning each critical region will contain 5 % of the distribution. Next, enter 0.05 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. Moreover, 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 B.
c Calculate the z-statistic using the following formula.
z = x-μ/ssqrt(n)
SubstituteValues
Substitute values
z = 25.5 - 24.3/5sqrt(65)
▼
Evaluate right-hand side
SubTerm
Subtract term
z = 1.2/5sqrt(65)
DivByFracD
a/b/c= a * c/b
z = 1.2 * sqrt(65)/5
UseCalc
Use a calculator
z=1.934941...
RoundDec
Round to 2 decimal place(s)
z ≈ 1.93
d See if the z-statistic falls within the critical region to make a conclusion about the initial claim. To do so, draw a point into the graph of the critical region to see where it falls.
Because the z-statistic falls in the critical region, the null hypothesis H_0 should be rejected. Given that the initial claim is related to the null hypothesis, there is strong evidence to reject the claim that the mean age of customers at the restaurant is 24.3. This means that it is more likely that the mean age is different than 24.3.
Confidence Intervals and Hypothesis Testing
Exercise 1.1
A survey of 150 randomly selected adults in California found that they spend an average of 45 minutes commuting to work every day with a standard deviation of 4.8 minutes. Use a 90 % confidence level to determine the maximum error of estimate for the amount of time spent commuting per day. Round to two decimal places.
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We will begin by recalling the formula for the maximum error of estimate. E=z*s/sqrt(n), n≥30 In this formula, z corresponds to the z-value of a particular confidence level, s is the sample standard deviation, and n is the sample size. To evaluate the formula, we first need to determine the value of z corresponding to the 90 % confidence level.
Finding the z-value
Because the confidence level 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.
Since the distribution is symmetric, the z-values limiting this area are opposites, so we only need to find one. Moreover, this value is given by the z-value of the upper or lower tail. We can determine this value by using a graphing calculator. We will push 2nd, then VARS, and choose the third option, invNorm(.
Next, we will enter 0.05 and push ENTER to get the z-value of the lower tail.
We found that the z-value is approximately -1.645, and because of the symmetry of the distribution, we can use the additive inverse 1.645 to evaluate the formula.
Evaluating the Formula
We have collected all the information to evaluate the formula of the maximum error of estimate. The standard deviation s is 4.8, the sample size n is 150, and the value of z is 1.645. Let's evaluate the formula.
Therefore, the maximum error of estimate at a 90 % confidence level is about 0.64.
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A factory that produces tennis balls measures the diameter of a sample of 35 balls.
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To determine which option is the right one, we need to use the given data to find the confidence interval for this situation. We will do this by following these steps.
- Identify the sample mean.
- Calculate the maximum error of estimate.
We are told that the sample mean consisting of 35 tennis balls was 6.7 centimeters. Next, we will calculate the maximum error of estimate. E=z*s/sqrt(n), n≥30 In this formula, z corresponds to the z-value of the confidence level, n is the sample size, and s is the sample standard deviation. Now, we need to determine the z-value corresponding to the 95 % confidence level. Let's look at the table showing the most commonly used confidence levels and their corresponding z-values.
| Confidence Level | z-value |
|---|---|
| 90 % | 1.645 |
| 95 % | 1.960 |
| 99 % | 2.576 |
From the table, we can see that the z-value corresponding to the 95 % confidence level is 1.960. We also have that the sample size n is 35 and the standard deviation s is 0.45. Let's substitute this information into the formula of the maximum error of estimate.
We have calculated the maximum error of estimate. We can now use the formula for the confidence interval of a population mean. CI = x ± E Recall that the sample mean x is 6.7. We will substitute this information in the previous formula to find the confidence interval.
| CI = x ± E | |
|---|---|
| x - E | x + E |
| 6.7 - 0.149 | 6.7 + 0.149 |
| 6.551 | 6.849 |
Therefore, the 95 % confidence interval is 6.551≤μ≤6.849.
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In the following situations, identify the claim to set the null and alternative hypotheses. Then select the options that appropriately describe these hypotheses.
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{"type":"multichoice","form":{"alts":["H_0: \u03bc= 336","H_a: \u03bc≠ 336","H_0: \u03bc\u2264 336","H_a: \u03bc>336"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":[0,1]}
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Let's start by identifying the claim of the tire manufacturer.
The tires we produce last at least 45 000 miles.
Let's represent this situation algebraically. We can do this with the following inequality. μ≥45 000miles This inequality represents the claim and because it is a non-strict inequality, it can be related to the null hypothesis H_0. Conversely, the complement of this claim is that the mean is less than 45 000, representing the alternative hypothesis H_a.
| Null Hypothesis | Alternative Hypothesis |
|---|---|
| The mean is greater than or equal to 45 000 miles. (claim) H_0:μ ≥ 45 000 |
The mean is less than 45 000. H_a:μ < 45 000 |
Following a similar process, we can determine the null and the alternative hypotheses for this situation. In this case, the coffee bags must weigh 336 grams. We can represent this situation as a statement of equality. μ=336grams In this case, the claim will represent the null hypothesis because of the equality sign. Conversely, the alternative hypothesis will be that the mean average of the coffee bags is not equal to 336 grams.
| Null Hypothesis | Alternative Hypothesis |
|---|---|
| The mean is equal to 336 grams. (claim) H_0:μ = 336 |
The mean is not equal to 336 grams. H_a:μ ≠ 336 |
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Ramsha and Tiffaniqua want to test whether a package distribution company can meet the promised delivery time of 3 days or less. They state the null and alternative hypotheses.
{"type":"choice","form":{"alts":["Ramsha","Tiffaniqua","Neither","Both"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":2}
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We will start by recalling that a hypothesis test is used to examine a claim about the mean of a population. The typical claims can be the following.
| 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. |
When performing a hypothesis test, we need to identify the null and alternative hypotheses.
- The null hypothesis H_0 is a statement of equality or non-strict inequality to be tested.
- The alternative hypothesis H_a is a statement of strict inequality that is the complement of the null hypothesis.
By writing the claim as an algebraic expression, we can identify whether it must be related to the null or the alternative hypotheses. With this in mind, let's consider the situation being analyzed by Ramsha and Tiffaniqua.
A package distribution company meets their promised delivery time of 3 days or less.
Note that the company claims that the delivery time is 3 days or less. This can be represented by the following non-strict inequality. μ≤ 3 days This claim will represent the null hypothesis because of the ≤ sign. Conversely, the alternative hypothesis will be the complement.
| Null Hypothesis | Alternative Hypothesis |
|---|---|
| The mean is less than or equal to 3 days. (claim) H_0:μ ≤ 3 |
The mean is greater than 3 days. H_a:μ > 3 |
Now let's review the hypotheses that Ramsha and Tiffaniqua wrote.
We can see that Tiffaniqua correctly relates the claim with the null hypothesis. However, she assigns the inequality signs wrongly. On the other hand, Ramsha identified the hypotheses correctly but misidentified the claim. Therefore, we can conclude that neither Ramsha nor Tiffaniqua is correct.
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Confidence Intervals and Hypothesis Testing
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