# A simple random sample of n=23 professors at a local community college have a mean course evaluation score of 3.91 with a standard deviation of 0.53. Assuming that the distribution of course evaluation scores is normally distributed, test the claim that the mean course evaluation score is 4.00 for all professors at this community college using a 0.05 significance level.

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A simple random sample of n=23 professors at a local community college have a mean course evaluation score of 3.91 with a standard deviation of 0.53. Assuming that the distribution of course evaluation scores is normally distributed, test the claim that the mean course evaluation score is 4.00 for all professors at this community college using a 0.05 significance level.

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Step 1

Solution:

Null and alternative hypotheses:

Null hypothesis: µ = 4

Alternative hypothesis: µ ≠ 4

Test statistic:

The population standard deviation is unknown and n < 30. Hence, the test statistic for this problem is shown below.

Step 2

Calculation of test statistic value:

Here, x-bar=3.91, s=0.53, n=23, µ = 4

Step 3

Degrees of freedom:

df = n – 1

= 23 – 1

= 22

Rejection region for two-tailed test:

Significance level is 0.05. the rejection region can be obtained using the EXCEL formula. “=T.INV.2T(0.05,22)”

Thus, the critical value of Student’s t is ttab= 2.07...

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