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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.
Answer:
Step-by-step explanation:
Universal set
U = 500
The number that likes snickers
n(S) = 42
The number that like Twix.
n(T) = 110
The number that like Reeses
n(R) = 125
n(S n T) = 33
n(T n R) = 62
n(S n R) = 26
n( S n R n T) = 22
Then,
n(S n T) only = n(S n T) - n(S n R n T)
n(S n T) only =33 - 22 = 11
n(T n R) only = n(T n R) - n(S n R n T)
n(T n R) only =62 - 22 = 40
n(S n R) only = n(S n R) - n(S n R n T)
n(S n R) only =26 - 22 = 4.
Also,
n(S) only = n(S) - n(S n R) - n(S n T) only
n(S) only = 42 - 26 - 11 = 5
n(T) only = n(T) - n(T n R) - n(T n S) only
n(T) only = 110 - 62 - 11 = 37
n(R) only = n(R) - n(S n R) - n(R n T) only
n(R) only = 124 - 26 - 40 = 58
Then, to know if some student don't like any of the of chocolate,
Let know the number of students that like the chocolate candy
n(S) + n(T)only + n(T n R)only + n(R) only
42 + 37 + 40 + 58 = 177 students.
Therefore, the total students that like chocolate candy is 177, so those that does not like any of them are
n(S U R U T)' = U – n(S U R U T)
n(S U R U T) = 500 - 177 = 323.
So, the question is how many student likes at most 2 kinds of these chocolates, this means that they can like exactly 2 or less or even none.
So, this category are
n(2 most) = n(S n T)only + n(R n T)only + n(S n R)only + n(s)only + n(T)only + n(R)only + n(S U R U T)'
n(2 most) = 11 + 40 + 4 + 5 + 37 + 58 + 323
n(2 most) = 478
The correct answer is E.
Check attachment for Venn diagram
