Question

1: Conduct the following hypothesis test. Specify the null and alternative hypotheses, and clearly state the conclusion of the test. Use the data from 6.3.3 (?̅ = 31.7, ? = 8.7, ? = 5). Is there enough evidence to conclude that the mean thymus weight is greater than 25mg? Set ? = 0.05

2: Suppose we used a two-tailed alternative for the previous test instead (thymus weight is not equal to 25mg). How would the p-value of the test change? What is the new p-value?

Answer 1

1) Given that, sample size (n) = 5, sample mean = 31.7 and

sample standard deviation (s) = 8.7

The null and alternative hypotheses are,

H0 : μ = 25

Ha : μ > 25

Test statistic is,

=> Test statistic = t = 1.722

Degrees of freedom = 5 - 1 = 4

Using Excel we find the p-value as shown below,

Excel Command : =TDIST (1.722, 4, 1) = 0.0801

=> P-value = 0.0801

Since, p-value is greater than significance level of 0.05, we should fail to reject the null hypothesis.

Conclusion : There is not sufficient evidence to conclude that the mean thymus weight is greater than 25mg.

2) Given that, the null and alternative hypotheses are,

H0 : μ = 25

Ha : μ ≠ 25

This test is two-tailed test.

We want to find the, p-value for this test, of test statistic = t = 1.722

Using Excel we find the p-value as shown below,

Excel Command : =TDIST (1.722, 4, 2) = 0.1602

=> P-value = 0.1602

OR

P-value = 2 * p-value of right-tailed test = 2 * 0.0801 = 0.1602

=> p-value = 0.1602

Since, p-value is greater than significance level of 0.05, we should fail to reject the null hypothesis.

Conclusion : There is not sufficient evidence to conclude that the mean thymus weight is not equal to 25mg.