MZYX=pi/3 radians. Covert this radian measure to its equivalent degree measure.
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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 = <u><em>population mean score</em></u>
So, Null Hypothesis, :
29.5 {means that the students that took the Kaplan tutoring have a mean score less than or equal to 29.5}
Alternate Hypothesis, :
> 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 <u>One-sample t-test statistics</u> because we don't know about population standard deviation;
T.S. = ~
where, = sample mean MCAT score = 32.2
s = sample standard deviation = 4.2
n = sample of students = 40
So, <u><em>the test statistics</em></u> = ~
= 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 <u><em>we have sufficient evidence to reject our null hypothesis</em></u> 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.