Question

 

1. In the past, the average teaching evaluation score at a college has been 4.10. The administration believes the move to distance learning has had a negative impact on teaching effectiveness . A random sample of 50 course sections has been selected. The sample showed an average teaching evaluation score of 4.03 and a standard deviation of 0.35. You want to test whether the administration's belief is legitimate. You want to test this hypothesis at 95% level of confidence. What is the p-values associated with this test? NOTE: WRITE YOUR ANSWER USING 4 DECIMAL DIGITS. DO NOT ROUND UP OR DOWN

2. A sample of 96 observations has been selected to test whether the population mean is greater than 25. The sample showed an average of 26 and a standard deviation of 5.7. You want to test this hypothesis at 90% level of confidence using the critical value approach. First, compute the critical value and the test statistics associated with this test. Second, compute the difference between the test statistic and the critical value (test statistic - critical value). What is this difference?NOTE: WRITE YOUR ANSWER WITH 4 DECIMAL DIGITS. DO NOT ROUND UP OR DOWN. MAKE SURE YOU INDICATE THE SIGN OF THIS DIFFERENCE CORRECTLY.

Answer 1

1.

HYPOTHESIS TEST-

Suppose, random variable X denotes teaching evaluation score in the college.

We have sample values. But we do not know population standard deviation (or variance). So, we have to perform one sample t-test.

We have to test for null hypothesis

against the alternative hypothesis

Our test statistic is given by

Here,

Sample size

Sample mean

Sample standard deviation

Degrees of freedom

[Using R-code 'pt(-1.414214,49)']

Level of significance

We reject our null hypothesis if

Here, we observe that

So, we cannot reject our null hypothesis.

Hence, based on the given data we can conclude that there is no significant evidence that the administration's belief is legitimate.

ANSWER-

The p-value associated with this test is 0.0818.

2.

HYPOTHESIS TEST-

Suppose, corresponding random variable be X.

We have sample values. But we do not know population standard deviation (or variance). So, we have to perform one sample t-test.

We have to test for null hypothesis

against the alternative hypothesis

Our test statistic is given by

Here,

Sample size

Sample mean

Sample standard deviation

Degrees of freedom

Level of significance

Critical value is given by [Using R-code 'qt(1-0.10,95)']

We reject our null hypothesis if

Here, we observe that

So, we reject our null hypothesis.

Hence, based on the given data we can conclude that there is significant evidence that population mean is greater than 25.

ANSWER-

Critical value is 1.2905.

Test statistic is 1.7189.

So, difference between test statistic and critical value = test statistic - critical value = 1.7189-1.2905 = 0.4284.