Respuesta :
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 [tex]\mu[/tex] = population mean score
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 29.5 {means that the students that took the Kaplan tutoring have a mean score less than or equal to 29.5}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 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. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean MCAT score = 32.2
s = sample standard deviation = 4.2
n = sample of students = 40
So, the test statistics = [tex]\frac{32.2-29.5}{\frac{4.2}{\sqrt{40} } }[/tex] ~ [tex]t_3_9[/tex]
= 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.
