Suppose you have been hired to study Guinea. You randomly sample 30 Guineans and

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

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