Question

3. A randomly selected sample of 14 students who stayed up all night to study for an exam received an average grade of 68% with a standard deviation of 8%. Another randomly selected sample of 12 students who wrote the same exam after a good night’s sleep received an average grade of 75% with a standard deviation of 7%. At a 5% significance level, does the sample provide enough statistical evidence to support a claim that students who stay awake all night perform worse on average? Conduct a 2-sample pooled variance t-test (i.e. t-test with equal variances) to evaluate the sample data.

A. State the null and alternative hypotheses – clearly indicate which is the null and which is the alternative. Use the symbol μA to represent the population mean grade for students who stayed awake and μS for students who slept in your hypotheses.

B. Calculate the degrees of freedom. Show all your work – no marks will be awarded without supporting calculations.

C. Look up the t critical value for this test and write out the decision rule using it.

D. Calculate the sample t statistic for this test. Show all your work – no marks will be awarded without supporting calculations.

E. In a full sentence answer: state whether or not you have rejected the null hypothesis AND respond to the original question (does the sample provide enough statistical evidence to support a claim that students

    who stay awake all night perform worse on average?)

Answer 1

For Awake :

x̅1 = 68, s1 = 8, n1 = 14

For Slept :  

x̅2 = 75, s2 = 7, n2 = 12

α = 0.05

A.

Null and Alternative hypothesis:

Ho : µA = µS

H1 : µA < µS

B.

df = n1+n2-2 = 24

C.

Critical value, t crit = T.INV(0.05, 24) = -1.711

Reject Ho if t < -1.711

D.

Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((14-1)*8² + (12-1)*7²) / (14+12-2) = 57.1250

Test statistic:

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (68 - 75) / √(57.125*(1/14 + 1/12)) = -2.3543

E.

Decision:

p-value < α, Reject the null hypothesis

Conclusion:

There is enough evidence to conclude that students who stay awake all night perform worse on average at 0.05 significance level.