A professor believes the students in a statistics class this term are more creative than most other students attending the university. A previous study found that students at the university had a mean score of 35 on a standard creativity test, and the current class has an average score of 40 on this scale with an estimated population standard deviation of 7. The standard deviation of the distribution of means is 1.63. What is the t score?

A. (40 – 35) / 7 = 0.71

B. (40 – 35) / 1.63 = 3.07

C. (40 – 35) / 7² = 5 / 49 = 0.10

D. (40 – 35) / 1.63² = 5 / 2.66 = 1.88

The professor can perform a One-mean t-test to determine whether the average score of the students in his class is more than the average score of all the students attending university.

A t-test will be used instead of the z-test because the population standard deviation is not provided instead it is estimated by the sample standard deviation.

The hypothesis for this test can be defined as follows:

H₀: The average score of the students in his class is not more then the entire university, i.e. μ ≤ 35.

Hₐ: The average score of the students in his class is more then the entire university, i.e. μ > 35.

Given:

[tex]\bar x=40\\SE_{\bar x}=1.63[/tex]

The test statistic is:

[tex]t=\frac{\bar x-\mu}{SE_{\bar x}}=\frac{40-35}{1.63}=3.0674\approx3.07[/tex]

Thus, the correct option is (B).