Question

A data set includes data from student evaluations of courses. The summary statistics are n=95, x=3.86, s=0.67. Use a 0.05 significance level to test the claim that the population of student course evaluations has a mean equal to 4.00. Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses?

Answer #1

Given:

n = 95

= 3.86

s = 0.67

**Null hypothesis:**

H0: The population mean,

**Alternative hypothesis:**

H1: The population mean,

**Level of Significance:**

**Test-statistic:**

**P-value:**

p-value is calculated using Excel with the function "TDIST". The inputs for the function are absolute value of t-statistic, degrees of freedom and number of tails. Absolute value of t-statistic is calculated as 2.0378, degrees of freedom = n - 1 = 95 - 1 = 94, number of tails = 2

Therefore, the final function is given as
**=TDIST(2.0378,94,2)**. p-value is calculated to be
0.0444

**Conclusion:**

p-value is 0.0444 which is less than the level of significance
of 0.05. Therefore, the null hypothesis is rejected and conclude
that the population mean of student course evaluation is
**not equal** to 4

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