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ALTA - CHAPTER 9 - HYPOTHESIS TESTING
EXAM 2025
Which of the following results in a null hypothesis p=0.61 and alternative hypothesis p>0.61? - Precise Answer ✔✔Correct answer: A study says that at most 61% of students study less than 5 hours per week. A researcher thinks this is incorrect, and wants to show that more than 61% of students study less than 5 hours per week. Remember that the alternative hypothesis is the claim that the researcher is trying to show. The null hypothesis, p=0.61 corresponds to the claim in the study, and the alternative hypothesis, p>0.61 corresponds to what the researcher is trying to show (to reject the null hypothesis). So the third choice is the correct answer Olivia, a golfer, claims that her drive distance is more than 174 meters, on average. Several of her friends do not believe her, so she decides to do a hypothesis test, at a 10% significance level, to persuade them. She hits 15 drives. The mean distance of the sample drives is 188 meters. Olivia knows from experience that the standard deviation for her drive distance is 14 meters. H0: μ=174; Ha: μ> α=0.1 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places? - Precise Answer ✔✔Correct answers:$
text{Test statistic = }3.87$Test statistic = 3.
The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean distance, x¯=188. The sample the golfer uses is 15 drives, so n=15. She knows the standard deviation of the drives, σ=14. Lastly, the golfer is comparing the population mean distance to 174 meters. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=188−1741415√≈143.615≈3. So, the test statistic for this hypothesis test is z0=3.87. Correct answers:$\text{Test statistic = }3.87$Test statistic = 3. The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean distance, x¯=188. The sample the golfer uses is 15 drives, so n=15. She knows the standard deviation of the drives, σ=14. Lastly, the golfer is comparing the population mean distance to 174 meters. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=188−1741415√≈143.615≈3. So, the test statistic for this hypothesis test is z0=3.87. - Precise Answer ✔✔Correct answers:$\text{Test statistic = }4.44$Test statistic = 4. The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean height, x¯=13.8. The sample the baker uses is 25 loaves, so n=25. She knows the standard deviation of the loaves, σ=0.9. Lastly, the baker is comparing the population mean height to 13 cm. So, this value (found in the null and alternative hypotheses) is
70% of its customers only shop at their sporting goods store when, in fact, it is at least 70%. Which of the hypothesis tests listed below is a left-tailed test? Select all correct answers. - Precise Answer ✔✔Correct answer: H0:μ=18, Ha:μ< H0:μ=11.3, Ha:μ<11. H0:μ=3.7, Ha:μ<3. Remember the forms of the hypothesis tests. Right-tailed: H0:μ=μ0, Ha:μ>μ0. Left-tailed: H0:μ=μ0, Ha:μ<μ0. Two-tailed: H0:μ=μ0, Ha:μ≠μ0. So in this case, the left-tailed tests are: H0:μ=3.7, Ha:μ<3. H0:μ=18, Ha:μ< H0:μ=11.3, Ha:μ<11. Suppose the null hypothesis, H0, is: the mean age of the horses on a ranch is 6 years. What is the Type II error in this scenario? - Precise Answer ✔✔Correct answer: You think the mean age of the horses on a ranch is 6 years when, in fact, it is not. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is thinking that the mean age of the horses is 6 years when, in fact, it is not
Which graph below corresponds to the following hypothesis test? - Precise Answer ✔✔The alternative hypothesis, Ha, tells us which area of the graph we are interested in. Because the alternative hypothesis is p>8.1, we are interested in the region greater than (to the right of) 8.1, so the correct graph is the second answer choice Which of the following results in a null hypothesis p=0.69 and alternative hypothesis p>0.69? - Precise Answer ✔✔Correct answer: A mechanic wants to show that more than 69% of car owners follow a normal maintenance schedule, contrary to a study that found that the percentage was at most 69%. Consider each of the options. Remember that the alternative hypothesis is the statement that the mechanic is trying to show. In this case, the mechanic wants to show that more than 69% of car owners follow a normal maintenance schedule, so the alternative hypothesis, Ha is p>0.69. This is shown in the second option. A mechanic wants to show that more than 44% of car owners do not follow a normal maintenance schedule. Identify the null hypothesis, H0, and the alternative hypothesis, Ha, in terms of the parameter p. - Precise Answer ✔✔Correct answer: H0: p=0.44; Ha: p>0. Let the parameter p be used to represent the proportion. Remember that the null hypothesis is the statement already believed to be true and the alternative hypothesis is the statement that is trying to be shown. In this case, the mechanic is trying to show that more than 44%
The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean score, x¯=155. The sample the bowler uses is 18 games, so n=18. She knows the standard deviation of the games, σ=17. Lastly, the bowler is comparing the population mean score to 140 points. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=155−1401718√≈154.007≈3. So, the test statistic for this hypothesis test is z0=3.74. Jolyn, a golfer, claims that her drive distance is not equal to 222 meters, on average. Several of her friends do not believe her, so she decides to do a hypothesis test, at a 5% significance level, to persuade them. She hits 11 drives. The mean distance of the sample drives is 218 meters. Jolyn knows from experience that the standard deviation for her drive distance is 14 meters. H0: μ=222; Ha: μ≠ α=0.05 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places? - Precise Answer ✔✔Correct answers:$
text{Test statistic = }-0.95$Test statistic = −0. The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean distance, x¯=218. The sample the golfer uses is 11 drives, so n=11. She knows the standard deviation of the drives, σ=14. Lastly, the golfer is comparing the population mean distance to 222 meters. So, this value (found in the null and alternative
hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=218−2221411√≈−44.221≈−0. So, the test statistic for this hypothesis test is z0=−0.95. Question Suppose a baker claims that his bread height is more than 15 cm, on average. Several of his customers do not believe him, so the baker decides to do a hypothesis test, at a 5% significance level, to persuade them. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm. The baker knows from baking hundreds of loaves that the standard deviation for the height is 0.5 cm. H0: μ=15; Ha: μ> α = 0.05 (significance level) What is the test statistic (zz-score) of this one-mean hypothesis test? - Precise Answer ✔✔Solution The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean height, x¯=17. The sample the baker uses is 10 loaves, so n=10. He knows the standard deviation of the height, σ=0.5. Lastly, the baker is comparing the population mean height to 15 cm. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=17−150.510√≈20.158≈12.
them. She throws 24 pitches. The mean speed of the sample pitches is 37 miles per hour. Floretta knows from experience that the standard deviation for her pitch speed is 5 miles per hour. H0: μ=46; Ha: μ< α=0.05 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places? - Precise Answer ✔✔Correct answers:$
text{Test statistic = }-8.82$Test statistic = −8. The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean speed, x¯=37. The sample the pitcher uses is 24 pitches, so n=24. She knows the standard deviation of the pitches, σ=5. Lastly, the pitcher is comparing the population mean speed to 46 miles per hour. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=37−46524√≈−91.021≈−8. So, the test statistic for this hypothesis test is z0=−8.82. Suppose a pitcher claims that his pitch speed is less than 43 miles per hour, on average. Several of his teammates do not believe him, so the pitcher decides to do a hypothesis test, at a 10% significance level, to persuade them. He throws 19 pitches. The mean speed of the sample pitches is 35 miles per hour. The pitcher knows from experience that the standard deviation for his pitch speed is 6 miles per hour. H0: μ=43; Ha: μ< α=0.1 (significance level)
What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places? - Precise Answer ✔✔Correct answers:$
text{Test statistic = }-5.81$Test statistic = −5. The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean speed, x¯=35. The sample the pitcher uses is 19 pitches, so n=19. He knows the standard deviation of the pitches, σ=6. Lastly, the pitcher is comparing the population mean speed to 43 miles per hour. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=35−43619√≈−81.376≈−5. So, the test statistic for this hypothesis test is z0=−5.81. Determine the Type II error if the null hypothesis, H0, is: a wooden ladder can withstand weights of 250 pounds and less. - Precise Answer ✔✔Correct answer: You do not reject the null hypothesis that the ladder can withstand weight of 250 pounds and less when, in fact, it cannot. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is stating that there is insufficient evidence to conclude that the ladder can withstand weight of 250 pounds, in fact, it cannot Suppose the null hypothesis, H0, is a surgical procedure is successful at least 80% of the time. And the alternative hypothesis, Ha, states the doctors' claim, which is a surgical procedure is successful less than 80% of the time.Which of the
A consumer protection company is testing a towel rack to see how much force it can hold. The null hypothesis, H0, is that the rack can hold 100 pounds of force. The alternative hypothesis, Ha, is that the rack can hold less than 100 pounds of force. What is a Type I error in this scenario? - Precise Answer ✔✔Correct answer: The researchers states that there is sufficient evidence to conclude that the rack holds less than 100 pounds of force, but the rack actually holds 100 pounds. Remember that a Type I error is rejecting the null hypothesis when the null hypothesis is true. A Type II error is not rejecting the null hypothesis when it is false. We are asked for the Type I error in this scenario. Rejecting the null hypothesis means rejecting the statement that the rack can hold 100 pounds. Therefore, a Type I error is: The researchers states that there is sufficient evidence to conclude that the rack holds less than 100 pounds of force, but the rack actually holds 100 pounds. Determine the Type I error if the null hypothesis, H0, is: a wooden ladder can withstand weights of 250 pounds and less. - Precise Answer ✔✔Correct answer: You think the ladder cannot withstand weight of 250 pounds and less when, in fact, it really can. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is thinking the wooden ladder cannot withstand the weights of 250 pounds or less when it really can.
Suppose the null hypothesis, H0, is: the mean age of the horses on a ranch is 6 years. What is the Type I error in this scenario? - Precise Answer ✔✔Correct answer: You cannot conclude that the mean age of the horses on a ranch is 6 years when, in fact, it is. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is rejecting the null hypothesis that the mean is 6 years when, in fact, it is. Determine the Type I error if the null hypothesis, H0, is: the percentage of homes in the city that are not up to the current electric codes is no more than 10%. And, the alternative hypothesis, Ha, is: the percentage of homes in the city that are not up to the current electric codes is more than 10%. - Precise Answer ✔✔Correct answer: There is sufficient evidence to conclude that more than 10% of homes in the city are not up to the current electrical codes when, in fact, there are no more than 10% that are not up to the current electric codes. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is when there is sufficient evidence to conclude that more than 10% of homes in the city are not up to the current electrical codes when, in fact, there are no more than 10% not up to the current electric codes. Suppose the null hypothesis, H0, is: no more than 70% of customers at a sporting goods store do not shop at any other sporting goods stores.And the alternative hypothesis, Ha, is: the sporting goods store claims more than 70% of its customers do not shop at any other sporting goods
Left-tailed: H0:p=p0, Ha:p<p0. Two-tailed: H0:p=p0, Ha:p≠p0. So, this is a two-tailed test. Identify the type of hypothesis test below.H0:X=16.9, Ha:X>16.9 - Precise Answer ✔✔Correct answer: The hypothesis test is right-tailed. Remember the forms of the hypothesis tests. Right-tailed: H0:X=X0, Ha:X>X0. Left-tailed: H0:X=X0, Ha:X<X0. Two-tailed: H0:X=X0, Ha:X≠X0. So, this is a right-tailed test. Determine the Type II error if the null hypothesis, H0, is: the mean price of a loaf of bread is $1.67. And the alternative hypothesis, Ha, states the claim, which is the mean price of a loaf of bread is not $1.67. - Precise Answer ✔✔Correct answer: You do not conclude that the mean price of a loaf of bread is not $1. when, in fact, the mean price of a loaf of bread is not $1.67. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is that you cannot conclude that the mean price of the bread is $1.67 when, in fact, it is not. Suppose the null hypothesis, H0, is a weightlifting bar can withstand weights of 800 pounds and less.And the alternative hypothesis, Ha, is a weightlifting bar can withstand weights of greater than 800 pounds.What
is α, the probability of a Type I error in this scenario? - Precise Answer ✔✔Correct answer: You conclude the weightlifting bar can withstand weights of greater than 800 pounds when, in fact, the weightlifting bar can withstand weights of less than or equal to 800 pounds. A consumer protection company is testing a seat belt to see how much force it can hold. The null hypothesis, H0, is that the seat belt can hold at least 5000 pounds of force. The alternative hypothesis, Ha, is that the seat belt can hold less than 5000 pounds of force. What is a Type II error in this scenario? - Precise Answer ✔✔Correct answer: The researchers states that there is insufficient evidence to conclude that the seat belt holds less than 5000 pounds of force, but the seat belt actually holds less than 5000 pounds. Remember that a Type I error is rejecting the null hypothesis when the null hypothesis is true. A Type II error is not rejecting the null hypothesis when it is false. We are asked for the Type II error in this scenario. Failing to reject the null hypothesis means failing to reject the statement that the seat belt can hold at least 5000 pounds. Therefore, a Type II error is: The researchers states that there is insufficient evidence to conclude that the seat belt holds less than 5000 pounds of force, but the seat belt actually holds less than 5000 pounds. Determine the Type I error if the null hypothesis, H0, is: 65% of college students will graduate with debt.And, the alternative hypothesis, Ha, is:
✔✔For a two-tailed hypothesis test, the rejection regions are on the left and the right. Because the test is two-tailed, there will be 2 critical values (one positive, one negative), which split the area under the normal curve into a middle 95%. We know that α=0.05, and so the area of the rejection region is equal to 0.05. But this is split between 2 areas, so instead of using zα, we need to use zα5. Since zα=z0.05, we know that: zα2=z0.052=z0. So the critical values are −z0.025=−1.960 and z0.025=1.960. Austin works in the marketing department for a large cable company. Recently, he heard that 56% of cable TV subscribers are considering dropping their cable TV subscription. He tests the claim by randomly selecting a sample of cable TV subscribers and asking whether they are considering dropping their cable TV subscription. Austin finds that 122 of the 224 he surveyed are considering dropping their cable TV subscription. What are the null and alternative hypotheses for this hypothesis test? - Precise Answer ✔✔{H0:p=0.56Ha:p≠0. First verify whether all of the conditions have been met. Let p be the population proportion for cable TV subscribers who are considering dropping their cable TV subscription. Since there are two independent outcomes for each trial, the proportion follows a binomial model. The question states that the sample was collected randomly. The expected number of successes, np=125.44, and the expected number of failures, nq=n(1−p)=98.56, are both greater than or equal to 5. Since Austin is looking for evidence that supports the claim that 56% of cable TV subscribers are considering dropping their cable TV subscription, the null hypothesis is that p is equal to 0.56 and the
alternative hypothesis is that p is not equal to 0.56. The null and alternative hypotheses are shown below. {H0:p=0.56Ha:p≠0. An airline company claims in a recent advertisement that more than 94% of passenger luggage that is lost is recovered and reunited with the customer within 1 day. Hunter is a graduate student studying statistics. For a research project, Hunter wants to find out whether there is convincing evidence in support of the airline company's claim. He randomly selects 315 passengers of the airline whose luggage was lost by the airline and found that 276 of those passengers were reunited with their luggage within 1 day. Are all of the conditions for this hypothesis test met, and if so, what are the null and alternative hypotheses for this hypothesis test? - Precise Answer ✔✔All of the conditions to conduct the hypothesis test have been met. The null and alternative hypotheses are {H0:p=0.94Ha:p>0.94. First verify whether all of the conditions have been met. Let p be the population proportion for the airline passengers whose luggage was lost by the airline and were reunited with their luggage within 1 day. Since Hunter is completing a survey where there are two independent outcomes, the proportion follows a binomial model. The question states that Hunter randomly selected the airline passengers whose luggage was lost by the airline. The expected number of successes, np=296.1, and the expected number of failures, nq=n(1−p)=18.9, are both greater than or equal to 5.
