Question

1. Suppose we take a sample from two separate populations and record some quantitative measurement for both. The first sample contained 60 respondents and resulted sample mean of 103 with a sample standard deviation of 8.2. The second sample contained 75 respondents and resulted sample mean of 100 with a sample standard deviation of 7.56. Using this information, our goal is to test:

H₀: μ1-μ2 = 0

H_a: μ1-μ2 > 0

What is the test statistic, t, for this example? Note: for your test statistic, keep the order of (sample #1) minus (sample #2).

1. 0.549
2. 1.980
3. 2.186
4. 3.165
5. 4.552

2. Using the information from the previous problem, which tail will you look at to find your p-value?

1. Left tail
2. Both tails
3. Right Tail

3. Refer to the information in the previous two questions. Using the df and t-distribution calculator Excel spreadsheet, how many degrees of freedom will this test use?

1. 78.445
2. 98.342
3. 99.987
4. 111.926
5. 121.666

4. Using the information from the previous three questions, what is the p-value for this hypothesis test?

1. 0.00098
2. 0.015
3. 0.019
4. 0.030
5. 0.067

5. Suppose we take sample from two separate populations and record some quantitative measurement for both. The results of these samples are given in the following table:

	ni $zBPIZkgqkYlajcAAAAASUVORK5CYII=$	yi $8EXmJpF+3KcV2YAAAAASUVORK5CYII=$	si $gJ6g1rtgAAAABJRU5ErkJggg==$
Sample #1	34	5.4	6.4
Sample #2	31	7.4	8.79

Use this information to test the following hypotheses:

H₀: μ1-μ2 = 0

H_a: μ1-μ2 < 0

What is the test statistic, t, for this example? Note: for your test statistic, keep the order of (sample #1) minus (sample #2).

1. The test statistic is less than -2.0.
2. The test statistic is between -2.0 and -1.5.
3. The test statistic is between -1.5 and -1.0.
4. The test statistic is between -1.0 and -0.5.
5. The test statistic is between -0.5 and 0.

6. Using the information from the previous problem, which tail will you look at to find your p-value?

1. Left tail
2. Both tails
3. Right tail

7. Refer to the information in the previous two questions. Using the df and t-distribution calculator Excel spreadsheet, how many degrees of freedom will this test use?

1. 31.879
2. 42.199
3. 54.446
4. 67.012
5. 88.430

Answer 1

1.
The provided sample means are shown below: X2 100 Also, the provided sample standard deviations are: 81-8.2 82 7.56 and the sample sizes are ni-60 and n275 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Testing for Equality of Variances A F-test is used to test for the equality of variances. The following F-ratio is obtained: 8 1,176 The critical values are FL = 0.982 and FU of equal variances is rejected. 1.013, and since F-1.176, then the null hypothesis (2) Rejection Region Based on the information provided, the significance level is 0.95, and the degrees of freedom are df = 121.666. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal: Hence, it is found that the critical value for this right-tailed test is te1.657, for a 0.95 and df 121.666. The rejection region for this right-tailed test is R t:t1.657

(3)_Test Statistics Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: X1 - X2 n1+n2-2 n2 103 100 (60-1)8.22+(75-1)7.562 ,1 2.186 60+75-2 4) Decision about the null hypothesis Since it is observed that t-2.186 > tcーー1.657, it is then concluded that the null hypothesis is rejected Using the P-value approach: The p-value is p concluded that the null hypothesis is rejected. 0.0154, and since p 0.0154 < 0.95, it is

(5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1 is greater than μ2, at the 0.95 significance level. Confidence Interval The 5% confidence interval is 2.914 < A-Ha < 3.086.

Graphically T-Test: t = 2.186, p = 0.0154 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 4.0-3.5 -3.0 -2.5-2.0 1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
1.(C)

2.(C)

3.(E)

4.(B)

5. same as 1st question calculation we get (C)

6.(A)

   7.(C)