Regina claims that the average score for her test 2 is less than 70%. You gather

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

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.