Q:

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 (dollars-squared). 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 against the alternative hypothesis that it has increased, at a 1% level of significance. What is the form of the rejection region that should be used to conduct this hypothesis test?

Accepted Solution

A:
Answer:Reject H0 if x > cvStep-by-step explanation:The hypothesis :H0 : μ ≤ 425H1 : μ > 425Standard deviation, s = √2500 = 50Sample size, n = 100xbar = 433.5The test statistic, Z :(xbar - μ) / s/√n(433.5 - 425) / 50/10Z = 8.5 / 5 = 1.7The decision region ;|Z| > Z0.01 ; Reject H0From Z table ;critical value, Z0.01 = 2.331.7 < 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.