Question

Question 3 (4 points) Does sleep deprivation decrease the speed at which people can complete mental tasks? Suppose a researcher knows that the average solving speed for a particular puzzle is u = 7.1 minutes with a standard deviation of o = 2 minutes. The researcher takes a random sample of 55 people and has each person stay up for 24 hours prior to completing the puzzle, then records the length of time each person takes to solve the puzzle. She wishes to use this data to test the hypotheses H H u u = 7.1 minutes > 7.1 minutes (a) [2 points] The researcher wants to conduct a test of the above hypotheses using a Type I error rate of 0.015. State the decision rule of this test in terms of the sample mean, Ē. (b) [1 point] Suppose this sample produced a sample mean of c = 7.56 minutes. Using the decision rule constructed in part (a), what decision do we come to with respect to our hypotheses? (c) [1 point) Find the probability of concluding that sleep deprivation does not decrease the speed at which people can complete mental tasks if the true average time in the population is u = 7.5 minutes.

Answer 1

Part a)

Reject null hypothesis if Z > Z(α)
Critical value Z(α) = Z(0.015) = 2.170

The values of sample mean X̅ for which null hypothesis is rejected
Z = ( X̅ - µ ) / ( σ / √(n))
2.17 = ( X̅ - 7.1 ) / ( 2 / √( 55 ))
X̅ >= 7.69

Part b)

Test Statistic :-
Z = ( X - µ ) / ( σ / √(n))
Z = ( 7.56 - 7.1 ) / ( 2 / √( 55 ))
Z = 1.7057

Test Criteria :-
Reject null hypothesis if Z > Z(α)
Critical value Z(α) = Z(0.015) = 2.17
Z < Z(α) = 1.7057 < 2.17
Result :- Fail to reject null hypothesis

Decision based on P value
Reject null hypothesis if P value < α = 0.015 level of significance
P value = P ( Z < 1.7057 )
P value = 0.044
Since 0.044 > 0.015 ,hence we reject null hypothesis
Result :- We fail to reject null hypothesis

There is insufficient evidence to support the claim that sleep deprivation decreases the speed at which people can complete mental task.

Part c)

The values of sample mean X̅ for which null hypothesis is rejected
Z = ( X̅ - µ ) / ( σ / √(n))
Critical value Z(α/2) = Z( 0.015 /2 ) = 2.17
2.17 = ( X̅ - 7.1 ) / ( 2 / √( 55 ))
Rejection region would be X̅ >= 7.69
P ( X̅ < 7.69 | µ = 7.5 ) = 0.7594
P ( type II error ) = ( ß ) = 0.7594