9.5d Conduct a Hypothesis Test for Mean - Population Standard Deviation Unknown P-Value Approach

Doctor Hector claims that the population mean birth weight of babies born at the hospital is exactly 8lb. Sheila, a nurse who assists Hector, wants to test this claim, so she takes a random sample of 55 babies born at the hospital. She determines that the sample mean weight is 7.8lb with a sample standard deviation of 0.9lb. Should Shelia reject or fail to reject the null hypothesis given the sample data below? H0:μ=8 versus Ha:μ≠8 α=0.05 (significance level) t0=−1.65 0.10<p-value<0.20

Fail to reject the null hypothesis because 0.10<p-value<0.20 is greater than the significance level α=0.05. 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 0.10<p-value<0.20 is greater than the significance level α=0.05, Shelia fails to reject the null hypothesis.

The USPS ships boxes that weigh exactly 2lb, according to their packaging. Elena, a USPS worker, tests the claim by hypothesizing that the population mean weight of the boxes is not equal to 2lb. She randomly selects 15 boxes and found that the sample mean weight is 1.9968lb with a standard deviation of 0.0286lb. Should Elena reject or fail to reject the null hypothesis given the sample data below? H0:μ=2 versus Ha:μ≠2 α=0.05 (significance level) t0=−0.43 p-value>0.20

Fail to reject the null hypothesis because p-value>0.20 is greater than the significance level α=0.05. 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 p-value>0.20 is greater than the significance level α=0.05, Elena fails to reject the null hypothesis.

Members of the school committee for a large city claim that the population mean class size of a middle school class within the district is exactly 20 students. Karla, the superintendent of schools for the city, thinks the population mean is less. She selects a random sample of 35 middle school classes across the city. The sample mean is 18.5 students with a sample standard deviation of 3.7 students. Should Karla reject or fail to reject the null hypothesis given the sample data below? H0:μ=20 versus Ha:μ<20 α=0.05 (significance level) t0=−2.40 0.01<p-value<0.025

Reject the null hypothesis because 0.01<p-value<0.025 is less than the significance level α=0.05. 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 0.01<p-value<0.025 is less than the significance level α=0.05, Karla rejects the null hypothesis.

Rebecca is a real estate agent who would like to find evidence supporting the claim that the population mean market value of houses in the neighborhood where she works is greater than $250,000. To test the claim, she randomly selects 35 houses in the neighborhood and finds that the sample mean market value is $259,860 with a sample standard deviation of $24,922. Should Rebecca reject or fail to reject the null hypothesis given the sample data below? H0:μ=250,000 versus Ha:μ>250,000 α=0.05 (significance level) t≈2.34 , which has 34 degrees of freedom 0.01<p-value<0.02

Reject the null hypothesis because 0.01<p-value<0.025 is less than the significance level α=0.05. 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 0.01<p-value<0.025 is less than the significance level α=0.05, Rebecca should reject the null hypothesis.

Kurtis is a statistician who claims that the average salary of an employee in the city of Yarmouth is no more than $55,000 per year. Gina, his colleague, believes this to be incorrect, so she randomly selects 61 employees who work in Yarmouth and records their annual salaries. The following is the data from this study: The alternative hypothesis Ha:μ>55,000. The sample mean salary of the 61 employees is $56,500 per year. The sample standard deviation is $3,750. The test statistic is calculated as 3.12. Using the information above and the portion of the t− table below, choose the correct p− value and interpretation for this hypothesis test.

The probability of observing a value of t0=3.12 or more if the null hypothesis is true is less than 0.005. Notice that the test statistic has 61−1=60 degrees of freedom and that this is a right-tailed test because the alternative hypothesis is Ha:μ>55,000. Find the p-value for right-tailed test of a t-distribution with 60 degrees of freedom, where t=3.12. Using the table of areas in the right tail for the t-distribution, in the row for 60 degrees of freedom, the t-test statistic, 3.12, is greater than 2.660. Since this is a right-tailed test, the p-value is less than 0.005.

A newspaper reports that the average annual salary of a full-time college professor in a certain region is $127,000. Scott, a statistics major, wants to test this claim. He selects a random sample of 75 full-time college professors in the region. The following is the data from this study: The alternative hypothesis Ha:μ≠127,000. The sample mean salary of the 75 college professors is $126,092. The sample standard deviation is $8,509. The test statistic is calculated as −0.92.

The probability of observing a value of t0=−0.92 or less or observing a value of t0=0.92 or more if the null hypothesis is true is greater than 0.20. Notice that the test statistic has 75−1=74 degrees of freedom and that this is a two-tailed test because the alternative hypothesis is Ha:μ≠127,000. Find the p-value for a two-tailed test of a t-distribution with 74 degrees of freedom, where t≈−0.92. Using the table of areas in the right tail for the t-distribution, in the row for 74 degrees of freedom, the absolute value of the t-test statistic, 0.92, is less than 1.293. Since this is a two-tailed test, the p−value is greater than 2∗0.10. Thus, p−value>0.20.

A car manufacturer claims that the company's newest model has a peak power of 329 horsepower. Katherine is a reporter for an automotive magazine who is testing the manufacturer's claim and says that the peak horsepower is less than the advertised amount. She measures the peak horsepower of 42 randomly selected cars of the new model. The following is the data from this study: The alternative hypothesis Ha:μ<329. The sample mean peak power of the 42 cars is 328.279 horsepower. The sample standard deviation is 2.41 horsepower. The test statistic is calculated as −1.94. Using the information above and the portion of the t− table below, choose the correct p− value and interpretation for this hypothesis test.

The probability of observing a value of t0=−1.94 or less if the null hypothesis is true is between 0.025 and 0.05. Notice that the number of degrees of freedom is 42−1=41 and that this is a left-tailed test because the alternative hypothesis is Ha:μ<329. Find the p-value for a left-tailed test of a t-distribution with 41 degrees of freedom, where t≈−1.94. Using the table of areas in the right tail for the t-distribution, in the row for 41 degrees of freedom, the absolute value of the t-test statistic, 1.94, is greater than 1.683 and less than 2.020. So, the p-value is between 0.025 and 0.05.

The mean body temperature of the human body under normal conditions is accepted to be 98.6∘F. Jason is a pre-med student who is skeptical of this claim. To test this claim, he selects a random sample of 47 people to record their body temperatures. The following is the data from this study: The alternative hypothesis Ha:μ≠98.6. The sample mean body temperature of the 47 people is 98.3∘F. The sample standard deviation is 0.80∘F. The test statistic is calculated as −2.57. Using the information above and the portion of the t− table below, choose the correct p− value and interpretation for this hypothesis test.

The probability of observing a value of t0=−2.57 or less or observing a value of t0=2.57 or more if the null hypothesis is true is between 0.01 and 0.02. Notice that the test statistic has 47−1=46 degrees of freedom and that this is a two-tailed test because the alternative hypothesis is Ha:μ≠98.6. Find the p-value for a two-tailed test of a t-distribution with 46 degrees of freedom, where t=−2.57. Using the table of areas in the right tail for the t-distribution, in the row for 46 degrees of freedom, the absolute value of the t-test statistic, 2.57, is greater than 2.410 and less than 2.687. Since this is a two-tailed test, the p−value is less than 2∗0.01 and greater than 2∗0.005. Thus, 0.01<p-value<0.02.

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

There is NOT sufficient evidence to conclude that the population mean weight of the boxes is not equal to 2lb. Compare the p-value to α=0.05. Since the p-value is greater than α, Elena fails to reject the null hypothesis H0. Therefore, there is NOT enough evidence at the α=0.05 level of significance to support the claim that the population mean weight of the boxes is not equal to 2lb.

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

There is NOT sufficient evidence to conclude that the population meanbirth weight of babies born at the hospital is not equal to 8lb. Compare the p-value to α=0.05. Since the p-value is greater than α, Shelia fails to reject the null hypothesis H0. Therefore, there is NOT enough evidence at the α=0.05 level of significance to support the claim that the population mean birth weight of babies born at the hospital is not equal to 8lb.