Suppose you have been hired to study Guinea. You randomly sample 30 Guineans and
ID: 3228058 • Letter: S
Question
Suppose you have been hired to study Guinea. You randomly sample 30 Guineans and find that that they make on average $970 and that 23 of them speak French. (Suppose that there is good reason to believe sigma = $75) a) You are asked to determine if Guineans make on average less than $1000. Test the hypothesis at the 5% significance level using the rejection region approach. b) Determine the p-value for the test in part (a). What can you conclude at the 1% significance level? c) For the test in part (a), determine the probability of making a type II error if the true mean is: i.) $950, ii.) $1250. d) For the test in part (a), determine the probability of making a type I error if the true mean is; i.) $1050, ii.) $950. e) Test the claim at the 0.5% significance level using z as the test statistic.Explanation / Answer
(a) Null HYpothesis : income >= $1000
Alternative hypothesis : income < $1000
Test Statistic: We will use t - test here
t = ( income - xincome bar)/ (s/n ) = ( 1000 - 970)/ ( 75/ 30) = 30/ 13.693 = 2.19
for dF = 29 and alpha = 0.05 => tcritical = 1.699
so t < tcritical so we can reject the null hypothesis
(b) so at alpha = 0.05 , P - vlaue = Pr ( X>= 1000; 970 ; 13.693) = 0.0143
P - value is less than the alpha level so we can reject the null hypothesis and say that mean income is less than $ 1000.
(c) True mean is $ 950 Then probability of making type II error is when we accept the null hypothesis when it is false.
so probability of making of type II error = Pr ( Mean income >= 1000; 950; 75)
Z = ( 1000 - 950)/ 75 = 0.67
so probability of making of type II error = 1 - (0.67) = 1- 0.7486 = 0.2514
True mean is $1250 Then probability of making of type II error is when we accept the null hypothesis when it is false.
so probability of making of type II error = Pr ( Mean income >= 1000; 950; 13.693)
Z = (1000 - 1250)/ 75 = -3.33
so probability of making of type II error = 0.0004
(D) making of type error when true mean = $1050
so type I error is to reject the null hypothesis when it is true
so Pr (type I error) = Pr ( X<= 1000; 1050; 13.693)
Z = ( 1000 - 1050)/13.693 = -3.65
Pr ( X<= 1000; 1050; 13.693) = (-3.65) = 0.0001
making of type error when true mean = $950
so type I error is to reject the null hypothesis when it is true
so Pr (type I error) = Pr ( X<= 1000; 950; 13.675)
Z = ( 1000 - 950)/13.675 =3.65
Pr ( X<= 1000; 950; 13.675) = (3.65) = 0.9999
(e) Test the claim at alpha = 0.5% = 0.005 significance level
Zcritical = 2.575
and z = 2.19 (as in part (a)
so we can not reject the null hypothesis and can say that the mean income level is equal or more tha