Question

1-2. A certain STAT 410 instructor claims that the time to complete a particular STAT 410 homework assignment is normally distributed with (unknown) mean u and standard deviation o= 7.2 minutes. A random sample of n= 81 students is obtained. 1. We wish to test Ho: u < 75 minutes vs. H1: u > 75 minutes. .. a) The average time to complete the homework assignment for the students in the sample was 76.2 minutes. Find the p-value. b) Find the Rejection Region for the test at a = 0.05. That is, for which values of the sample mean X should we reject Ho, if a 5% level of significance is used? c) Find the power of the test if the true value of the average time to complete the assignment is (i) 75.7 minutes and (ii) 76.7 minutes, and a 5% level of significance is used. d) What is the minimum sample size required if we want to have the power of at least 2/3 at u = 75.7 minutes? Use a = 0.05.

Answer 1

1-2 Let arce. x denotes the line to coupheld a particular Stat 410 bomework assiqmuerend Given, x~ N(460) where Al is unknown and 8=7.9 dét a random sample of n = 01 students is obtained. Let x be the sa sample We want to test, mean . Hoi esys vs Hif775 the to, the critical region is as C= { For rejecting * >K?

Z= @ Given, F = 76.2 Then, the lesp statistica x - El ~ N(0,1) Con 76.2-75 under to 7.21581 1.289 7.2 11 ote mpor = -1.5 Then, p-value = P(2715) = 0.0660 (Deuce. p-value = 0.0668

b Ho Given, x=0.05 bliana X= p (&>k (Ho) 30.08=p(x-le > ohn 0150 50.05 =P(27 K-75 72/18/ 20-05 = P(2> 9CK-75)) q(K-75) 20.95 7.2 7.2 K-76 = Z-ax 1.6449 = K = 75+ (7:2x1.6449) K = 75+ 1.3159 .: K= 76.3159 Meneer Tu rejection reqion as C: {x: 8776.3898

Let t7 be Ma power of die Ceste Let = 75.7 Then T = P(8 776 3/59/H) EPCX-4376-389-42/H) olun มา = P(Z > 76359.757 7.21561 = P(27 0-8159x9 6/ ) =P(21-0199) - 0.1859 Moneer power. Bf the list al 0 - 04397

2 р (14) 02/2) (11 If ce= 76.7 than T: PC*> 76-3/89/H) X-Cl > 26.3159 ke own =P(2>7387-767) 26.3159-76-7 = P(27-0.384109 =plzz – 0.4801) = 0.8641 Heveer power of léste caso 1 = 0.0641 72

a det sample size ke n and given powsze the 17 - 2 3 75.7 and re = 78.7 Then te p (*776.3159 (Hi) 76.3159.4%. > 0 > este olu : 2 = P(27676-3159-75-7) 72(du 3 76.3189-75:7) Z1 / E 0.615 9 in ²0.4307 7.2

Jn 0.4307878 0.6159 > In 5.0407 n = 25.6087 -> ! nu 25 tence, required sample size as n=85.