Regina claims that the average score for her test 2 is less than 70%. You gather
ID: 3265501 • Letter: R
Question
Regina claims that the average score for her test 2 is less than 70%. You gather a sample of 9 tests and calculate the sample average to be 64% and the population standard deviation to be 27.6%.
a) Write out the description of the hypothesis test using the standard notation and state whether it is one-sided or two-sided.
b) Test the hypothesis using the P-Value approach. What does the computed P-value suggest about the hypothesis?
c) Test the hypothesis using the critical region (fixed interval) approach. Assume = 0.04. What is / are the value(s) of the boundaries of the fixed interval? What does this test suggest about the hypothesis?
d) Suppose Aimee Birdsall knows the true test average is 77%. What is the probability of a Type II Error under these circumstances?
e) What is the power of this test? How many samples would you need to get the power of the test to 95%?
f) Compute the 95% confidence intervals for the mean value of the sample. What do the confidence intervals suggest about the hypothesis?
Explanation / Answer
Question a)
H0: µ 70%
H1: µ < 70%
This is one-tailed test.
Question b)
Test statistics:
Z = ( x bar - µ)/(s/sqrt(n))
= (64 – 70)/(27.6/sqrt(9))
= -0.65
p-value = P (z<-0.65)
By using normal table we get P ( z < -0.65) as 0.2578.
The p-value is 0.2578
Level of significance is .04
The p-value is greater than the level of significance we fail to reject the null hypothesis.
There is no sufficient evidence to conclude that the average score for her test 2 is less than 70%.
Question c)
The critical z at .04 level of significance is given as -1.75. Here the test statistics value is greater than the critical value (-0.65 >-1.75) we fail to reject the null hypothesis.
There is no sufficient evidence to conclude that the average score for her test 2 is less than 70%.
Question d)
z = x bar + (z* SE)
=70-(1.75*(27.6/SQRT(9)))
= 53.9
P ( x bar > 53.9)
z = (53.9 – 77)/(27.6/sqrt(9))
= -2.51
P ( z > -2.51)
By using normal table we get P ( z >-2.51) is 0.9940.
Probability of Type II error is 0.9940