Question

The mean MCAT score 29.5. Suppose that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.2 with a standard deviation of 4.2. Test the claim that the students that took the Kaplan tutoring have a mean score greater than 29.5 at a 0.05 level of significance.

Answer 1

Answer:

We conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

Step-by-step explanation:

We are given that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.2 with a standard deviation of 4.2.

Let $\mu$ = population mean score

So, Null Hypothesis, $H_0$ : $\mu \leq$ 29.5      {means that the students that took the Kaplan tutoring have a mean score less than or equal to 29.5}

Alternate Hypothesis, $H_A$ : $\mu$ > 29.5      {means that the students that took the Kaplan tutoring have a mean score greater than 29.5}

The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;

                               T.S.  =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean MCAT score = 32.2

            s = sample standard deviation = 4.2

            n = sample of students = 40

So, the test statistics =  $\frac{32.2-29.5}{\frac{4.2}{\sqrt{40} } }$  ~  $t_3_9$

                                    =  4.066

The value of t-test statistics is 4.066.

Now, at 0.05 level of significance, the t table gives a critical value of 1.685 at 39 degrees of freedom for the right-tailed test.

Since the value of our test statistics is more than the critical value of t as 4.066 > 1.685, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.