Question

A development economist is studying income growth in a rural area of a developing country. The last census of the population of this area, several years earlier, showed that mean household annual income was 425 dollars, and the variance of household income was 2500 (dollarssquared). A current random sample of 100 households yields a sample mean income of $433.75. Assume that household annual incomes are approximately normally distributed, and that the population variance is known still to be 2500. Test the null hypothesis that population mean income has not increased ( H0 : "m"(greek letter) greater than equal to< 425 ) against the alternative hypothesis that it has increased ( HA :"m"(greek letter > 425 ), at a 1% level of significance.
What is the form of the rejection region?
a. Reject H0 if x < cv
b. Reject H0 if x > cv
c. Reject H0 if x < cv or if x > cv
d. Reject H0 if x = cv 3
What is the critical value?
a. 436.63
b. 433.75
c. 425
d. 437.88
What is the conclusion of the hypothesis test?
a. Accept H0
b. Reject H0 in favor of HA

Answer 1

Answer:

Reject H0 if x > cv

Step-by-step explanation:

The hypothesis :

H0 : μ ≤ 425

H1 : μ > 425

Standard deviation, s = √2500 = 50

Sample size, n = 100

xbar = 433.5

The test statistic, Z :

(xbar - μ) / s/√n

(433.5 - 425) / 50/10

Z = 8.5 / 5 = 1.7

The decision region ;

|Z| > Z0.01 ; Reject H0

From Z table ;critical value, Z0.01 = 2.33

1.7 < 2.33 ; We fail to reject then Null and conclude that thee is no significant evidence support the claim that population mean income has increased.