Answers and explanations to all numbered questions
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Past records suggest that the mean annual income, mu1, of teachers in state of Georgia is greater than or equal to the mean annual income, mu2, of teachers in Indiana. In a current study, a random sample of 20 teachers from Georgia and an independent random sample of 20 teachers from Indiana have been asked to report their mean annual income. The data obtained are as follows.

Annual income in dollars
Georgia 30964,36499,43508,40036,29752,36613,35317,42073,42721,37973,37954, g31990,42717,46182,35264,43142,43170,33550,38698,31504
Indiana 40452,39088,38551,35948,49637,40013,5215034665,41984,43107,37433,
i42592,46894,33281,45760,49092,40188,40973,33208,46104

The population standard deviation for mean annual income of teachers in Georgia and in Indiana are estimated as 6500 and 6600, respectively. It is also known that both populations are approximately normally distributed. At the 0.01 level of significance, is there sufficient evidence to reject the claim that the mean annual income of teachers in state of Georgia is greater than or equal to the mean annual income of teachers in Indiana? Perform a one-tailed test. Provide answers to the questions below.

Carry your intermediate computations to at least three decimal places and round your answers as specified below.

I've summarized the information from the problem (I calculated the mean from the data, the sample size and standard deviations are given to us):

Georgia: Indiana:
n = 20 n = 20
mean = 37981.35000 mean = 41556.00000
sd = 6500 sd = 6600

The population standard deviations are given to us.

Do one-tailed test with significance level of 0.01.

1. The null hypothesis Ho =

The mean annual income of teachers in state of Georgia is greater than or equal to the mean annual income of teachers in Indiana.

μ1 ≥ μ2

2. The alternative hypothesis H1 =

The mean annual income of teachers in state of Georgia is less than the mean annual income of teachers in Indiana.

&#956;1 < &#956;2

3. The type of test statistic(choose one) Z , t if so, provide degrees of freedom , chi square if so, provide degrees of freedom , F if so provide dfn and dfd

We're going to use the z statistic. To compare the means of two groups, you use either the z or the t statistics. If you don't know the population standard deviation (you only know the sample standard deviation), you use t. In this case, since the question tells us what the population standard deviations are, we can use z.

4. The value of the test statistic (round to at least 3 decimal places)

We're going to use a two-sample z-test. The formula for that is:

So, to find z, we subtract the second sample mean from the first, and divide by the standard deviation of the difference. The standard deviation of the difference is the square root of the squared standard deviation of the first divided by the sample size plus the squared standard deviation of the second divided by the sample size.

z = 37981.35 - 41556.00 = -3574.65/2071.35 = -1.72576
&#8730;(65002/20 + 66002/20)

5. The critical value at the 0.01 level of significance (round to at least 3 decimal places)

If you look at a z-distribution table, you'll see that the critical value for z for a one-tailed test at the 0.01 level is z = 2.326. Because we're testing the hypothesis &#956;1 < &#956;2 (one mean is less than the other), we will use the negative value. So the critical value is z = -2.326.

6. Can we reject the claim that the mean annual income of teachers from Georgia is greater than or equal to the mean annual income of teachers from Indiana? Yes or No

We can reject the null hypothesis (and thereby reject the claim that the mean annual income of teachers in state of Georgia is greater than or equal to the mean annual income of teachers in Indiana) only if our value of z is less than -2.326. We cannot do that here because -1.726 is NOT less than -2.326. Therefore we CANNOT reject the null hypothesis.

please see the attached file.