9.5b Conduct a Hypothesis Test for Mean - Population Standard Deviation Known P-Value Approach

Bobbie is interested in buying a new home, and would like to determine if housing prices have gone up this year. After determining the mean and standard deviation for housing costs, she calculated a test statistic of z0=2.05. Using the alternative hypothesis Ha:μ>150,000, what is the p-value of a right-tailed one-mean hypothesis test, with a test statistic of z0=2.05? (Do not round your answer; compute your answer using a value from the table below.)

The probability of observing a value of z0=2.05 or more if the null hypothesis is true is 2%. The p-value is the probability of an observed value of z=2.05 or greater if the null hypothesis is true, because this hypothesis test is right-tailed. This probability is equal to the area under the Standard Normal curve to the right of z=2.05. Using the Standard Normal Table, we can see that the p-value is equal to 0.98, which is the area to the left of z=2.05. (Standard Normal Tables give areas to the left.) So, the p-value we're looking for is p=1−0.98=0.02.

Using the information above, choose the correct conclusion for this hypothesis test.

There is NOT sufficient evidence to conclude that the average amount residents spend per month on public transit fare is less than $140. Since Rachel is making the claim that the average amount that residents spend per month on public transit fare is less than $140, this is a left-tailed distribution. If the decision is to reject the null hypothesis, then we can conclude there is enough evidence to support the claim. However, since the p-value 0.3015 is greater than the significance level α=0.05, do not reject the null hypothesis. Rachel does not have sufficient evidence to conclude that the average amount spent per month on public transit fare is less than $140.

Using the information above, choose the correct conclusion for this hypothesis test.

There is NOT sufficient evidence to conclude that the headache medication gives patients relief in less than 61 minutes, on average. Since Jacqueline is making the claim that the headache medication gives patients relief in 61 minutes or less, on average, this is a left-tailed distribution. If the decision is to reject the null hypothesis, then we can conclude there is enough evidence to support the claim. However, since the p-value 0.0233 is greater than the significance level α=0.01, do not reject the null hypothesis. Jacqueline does not have sufficient evidence to conclude that the headache medication gives patients relief in less than 61 minutes, on average.

What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1.01? (Do not round your answer; compute your answer using a value from the table below.)

0.312 The p-value is the probability of an observed value of z=1.01 or greater in magnitude if the null hypothesis is true, because this hypothesis test is two-tailed. This means that the p-value could be less than z=−1.01, or greater than z=1.01. This probability is equal to the area under the Standard Normal curve that lies either to the left of z=−1.01, or to the right of z=1.01. Using the Standard Normal Table given, we can see that the p-value that corresponds with z=−1.01 is 0.156, which is just the area to the left of z=−1.01. Since the Standard Normal curve is symmetric, the area to the right of z=1.01 is 0.156 as well. So, the p-value of this two-tailed one-mean hypothesis test is (2)(0.156)=0.312.

What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1? (Do not round your answer; compute your answer using a value from the table below.)

0.318 The p-value is the probability of an observed value of z=1 or greater in magnitude if the null hypothesis is true, because this hypothesis test is two-tailed. This means that the p-value could be less than z=−1, or greater than z=1. This probability is equal to the area under the Standard Normal curve that lies either to the left of z=−1, or to the right of z=1. Using the Standard Normal Table given, we can see that the p-value that corresponds with z=−1 is 0.159, which is just the area to the left of z=−1. Since the Standard Normal curve is symmetric, the area to the right of z=1 is 0.159 as well. So, the p-value of this two-tailed one-mean hypothesis test is (2)(0.159)=0.318.

What is the p-value of a right-tailed one-mean hypothesis test, with a test statistic of z0=0.36? (Do not round your answer; compute your answer using a value from the table below.)

0.359 The p-value is the probability of an observed value of z=0.36 or greater if the null hypothesis is true, because this hypothesis test is right-tailed. This probability is equal to the area under the Standard Normal curve to the right of z=0.36. Using the Standard Normal Table, we can see that the p-value is equal to 0.641, which is the area to the left of z=0.36. (Standard Normal Tables give areas to the left.) So, the p-value we're looking for is p=1−0.641=0.359.

Jacqueline, a medical researcher, would like to make the claim that a newly developed prescription medication for headaches gives patients relief in less than 61 minutes, on average. Jacqueline samples 12 users of the medication and obtains a sample mean of 55 minutes. At the 1% significance level, should Jacqueline reject or fail to reject the null hypothesis given the sample data below? H0:μ=61 minutes; Ha:μ<61 minutes α=0.01 (significance level) test statistic=−1.99 Use the graph below to select the type of test (left-, right-, or two-tailed).

Do not reject the null hypothesis because the p-value 0.0233 is greater than the significance level α=0.01. Jacqueline is making the claim that the average time it takes the headache medication to work is less than 61 minutes. Therefore, this is a left-tailed test because of "less than". Since this is a left-tailed test, α and p-value will be areas on the left side of the distribution. In making the decision to reject or not reject H0, recall the criterion for comparing the p−value to α. If p≤α, reject H0. Since the p-value 0.0233 is greater than the significance level α=0.01, do not reject the null hypothesis.

Rachel, a city employee, would like to make the claim that the average amount that residents spend per month on public transit fare is less than $140. Rachel samples 25 city residents and obtains a sample mean of $125.80 spent per month on public transit. At the 5% significance level, should Rachel reject or fail to reject the null hypothesis given the sample data below? H0:μ=$140 per month; Ha:μ<$140 per month α=0.05 (significance level) test statistic=−0.52 Use the graph below to select the type of test (left-, right-, or two-tailed). Then set the α and the test statistic to determine the p-value. Use the results to determine whether to reject or fail to reject the null hypothesis.

Do not reject the null hypothesis because the p-value 0.3015 is greater than the significance level α=0.05. Rachel is making the claim that the average amount that residents spend per month on public transit fare is less than $140.Therefore, this is a left-tailed test because of "less than." Since this is a left-tailed test, α and p-value will be areas on the left side of the distribution. In making the decision to reject or not reject H0, recall the criterion for comparing the p−value to α. If p≤α, reject H0. Since the p-value 0.3015 is greater than the significance level α=0.05, do not reject the null hypothesis.