The Student's t distribution table gives critical values for the Student's t distribution. Use an appropriate d.f. as the row header. For a right-tailed test, the column header is the value of α found in the one-tail area row. For a left-tailed test, the column header is the value of α found in the one-tail area row, but you must change the sign of the critical value t to −t. For a two-tailed test, the column header is the value of α from the two-tail area row. The critical values are the ±t values shown.Let x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the x distribution is μ = 7.4†. A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 41 patients with arthritis took the drug for 3 months. Blood tests showed that x = 8.0 with sample standard deviation s = 1.8. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood. Solve the problem using the critical region method of testing (i.e., traditional method). (Round your answers to three decimal places.) test statistic =   critical value = ±   State your conclusion ion the context of the application. Reject the null hypothesis, there is insufficient evidence that the drug has changed the mean pH level.Fail to reject the null hypothesis, there is insufficient evidence that the drug has changed the mean pH level.    Reject the null hypothesis, there is sufficient evidence that the drug has changed the mean pH level.Fail to reject the null hypothesis, there is sufficient evidence that the drug has changed the mean pH level. Compare your conclusion with the conclusion obtained by using the P-value method. Are they the same? We reject the null hypothesis using the traditional method, but fail to reject using the P-value method.The conclusions obtained by using both methods are the same.    We reject the null hypothesis using the P-value method, but fail to reject using the traditional method.

A First Course in Probability (10th Edition)
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ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
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The Student's t distribution table gives critical values for the Student's t distribution. Use an appropriate d.f. as the row header. For a right-tailed test, the column header is the value of α found in the one-tail area row. For a left-tailed test, the column header is the value of α found in the one-tail area row, but you must change the sign of the critical value t to −t. For a two-tailed test, the column header is the value of α from the two-tail area row. The critical values are the ±t values shown.

Let x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the x distribution is μ = 7.4†. A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 41 patients with arthritis took the drug for 3 months. Blood tests showed that x = 8.0 with sample standard deviation s = 1.8. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood. Solve the problem using the critical region method of testing (i.e., traditional method). (Round your answers to three decimal places.)

test statistic =  
critical value = ±  


State your conclusion ion the context of the application.

Reject the null hypothesis, there is insufficient evidence that the drug has changed the mean pH level.Fail to reject the null hypothesis, there is insufficient evidence that the drug has changed the mean pH level.    Reject the null hypothesis, there is sufficient evidence that the drug has changed the mean pH level.Fail to reject the null hypothesis, there is sufficient evidence that the drug has changed the mean pH level.


Compare your conclusion with the conclusion obtained by using the P-value method. Are they the same?

We reject the null hypothesis using the traditional method, but fail to reject using the P-value method.The conclusions obtained by using both methods are the same.    We reject the null hypothesis using the P-value method, but fail to reject using the traditional method.
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