Question

A researcher believes that community B has a lower income than community A. Forty people are surveyed in each of the two communities regarding their income. In place A, the mean among the 40 individuals is 62 (given in thousands; you can ignore the ‘thousands’ and just use ‘61’), and the variance is 95; in place B, the mean among the 40 surveyed is 59, and the variance is 100.   Test the null hypothesis that the two communities have equal incomes. Use alpha = 0.05. State the null and alternative hypotheses, find the test statistic, compare it with the critical value, make a decision, and find the p-value.

Thank you!

Answer 1

Null and alternative hypothesis is

Ho: mu1 = mu2

Ha: mu1>mu2

Where mu1 and mu2 are average income of community A and community B respectively

Test statistic

Z = (x1bar-x2bar) / sqrt(sigma1^2/n1 + sigma2^2/n2)

Where x1bar and x2bar are sample average income of community A and B. n1 and n1 are respective sample size

Z = (62-59) / sqrt(95^2/40 + 100^2/40)

= 3/21.81

= 0.137

Critical value at 5% of one tailed z test is 1.645. Since test statistic is less than 1.645, we fail to reject Ho at 5%. Hence researcher claim is not correct. P value corresponding to one tailed z test with test statistic z = 0.137 is 0.445