A random sample was taken of 64 students who took the ACT after comp leting a ce

Question

A random sample was taken of 64 students who took the ACT after comp leting a certain prep course for the ACT . The sample mean test score is 23 and the sample standard deviation of the test scores is 2.

a. Which distribution you will use and why?

b. Calculate a 9 2 % confidence interval for the populatio n mean test score of students taking the prep course.

c. Administrators of the course are interested determining whether the average ACT score for students taking the prep course is greater than 21.1 (the average of all students taking the ACT that year). Conduct a hypothesis te st using a significance level of =0.05 to determine if there is convincing evidence to support this.

Explanation / Answer

a.
T-distribution
b.
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=23
Standard deviation( sd )=2
Sample Size(n)=64
Confidence Interval = [ 23 ± t a/2 ( 2/ Sqrt ( 64) ) ]
= [ 23 - 1.779 * (0.25) , 23 + 1.779 * (0.25) ]
= [ 22.555,23.445 ]
c.
Set Up Hypothesis
Null, H0: U=21.1
Alternate,students taking the prep course is greater than 21.1 H1: U>21.1
Test Statistic
Population Mean(U)=21.1
Sample X(Mean)=23
Standard Deviation(S.D)=2
Number (n)=64
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =23-21.1/(2/Sqrt(64))
to =7.6
| to | =7.6
Critical Value
The Value of |t | with n-1 = 63 d.f is 1.669
We got |to| =7.6 & | t | =1.669
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Right Tail - Ha : ( P > 7.6 ) = 0
Hence Value of P0.05 > 0,Here we Reject Ho

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