Question

Please use a 1-sample test hre. There is only a single random sample. Thanks in advance!

A sample of 618 students with learning disability who received accommodations at the university level was taken. Those students earned an average ACT score of 25.9. Assume a known standard deviation ??= 5.4. The average ACT score for students with learning disability who do not receive accommodations is 26.4. Test whether the average ACT score for students who receive accommodations is different from the ACT score of those without accommodations. Assume ??=0.01.
a. Label and describe the parameters involved.
b. State the hypotheses.
c. State the significance level.
d. Calculate the test statistic.
e. What is the critical value? Show Excel formula used or reference ICON table.
f. What is the p-value? Show Excel formula used or reference ICON table.
g. What is the two-sided 99% confidence interval for the true mean ACT score for students with a learning disability who receive accommodations?
h. What are your conclusions about the test? Interpret the result in the context of the problem.

Answer 1

a) Here population standard deviation is known so we have to use z-test statistic. Here population mean is a parameter.

ACT score of students is a parameter.

b) Claim: The ACT score for the students who receive accommodation is different from the ACT score of those without accommodation.

The null and alternative hypothesis is

c) Level of significance = 0.01

d)

Sample size = n = 618

Sample mean = = 25.9

Population standard deviation = = 5.4

Test statistic is

d) Level of significance = 0.01

Critical value = 2.58 ( Excel formula =NORMSINV(0.995))

Critical value > Test statistic we fail to reject null hypothesis.

f) P-value = 2*P(Z < - 2.30) = 2*0.0107 = 0.0213 ( Excel formula =NORMSDIST(-2.30)

P-value > 0.01 we fail to reject null hypothesis.