Question

International students from Asian countries often have trouble adjusting to American classrooms. Researchers hypothesize that Asian students might talk less often. A study evaluated the propensity of Asian males to speak in class. On the scale in the study, American males have a mean of 10.0. A random sample of 64 Asian males had a sample mean of 8.75 and sample standard deviation of 2.57. Conduct a hypothesis test at α = 0.05 to see if Asian males speak less often than the population of American males .A. [3] What are the null and alternative hypotheses? B. [3] Are the conditions for a hypothesis test satisfied? Why or why not? C. [3] What is the rejection region?D. [5] Compute the test value. E. [3] Make a decision and briefly explain why you made your decision. F. [3] Write a sentence explaining your decision in context.

Answer 1

Objective: To test whether the Asian males speak less often than the population of American males.Let denote the population mean measure of propensity of Asian males to speak in class. We have to compare this with the mean of the measure of propensity of American males to speak in class, which is given to be equal to 10.

Given: Sample Mean , Sample standard deviation s = 2.57, Sample size n = 64

Hence, the Null and Alternative Hypothesis can be expressed as follows:

Vs at  

As mentioned, since, the population standard deviation is unknown, the appropriate test to test the above hypothesis would be a one sample t test:  

But before running this test, we must ensure that the data satisfies the assumptions of this test:

- The data is continuous - The observations are independent - The data is normally distributed - There are no outliers   

Assuming that all the assumptions are satisfied:

The test statistic is given by:

with critical region given by: for left tailed test.

From the given data,

Substituting the values obtained in the test statistic:

= -3.891

Comparing this observed test statistic value with the critical value

Using excel,

..............(since, it is a one tailed test and excel function is only for two tailed test; hence 2*0.05 = 0.10; Here, since this is a left tailed test, the t value would have a negative sign)

We get

Since, t =-3.891 < -1.666, does lie in the critical region, we may reject H0 at 5% level.

Also, the p-value of the test is obtained as:

We get p-value = 0.0001 < 0.05. The probability of obtaining an extreme result, as the one obtained, when H0 is true, is very low, there is chance that H0, might not be true. Hence, we may reject H0 at 5% level. We may conclude that the data provides sufficient evidence to support the claim that the Asian males speak less often than the population of American males.