Question

A.

% 9.4.10-T Question Help The accompanying table gives results from a study of words spoken in a day by men and women. Assume that both samples are independent simple random samples from populations having normal distributions. Use a 0.05 significance level to test the claim that the numbers of words spoken in a day by men vary more than the numbers of words spoken in a day by women. Men Women 185 211 15.668.5 16,215. 9 8,632.7 ,301.1 7 What are the null and alternative hypotheses? O A. Ho: 0% = 0 H:03+03 OC. He: 042 H7:02 = 0 B. He: 0 = 0 - H4:0? > 0 OD. Hai c?ap? H4:0? <a Identify the test statistic. F=L(Round to two decimal places as needed.) Enter your answer in the answer box and then click Check Answer. parts 2 remaining Clear All Check Answer

B.

2. Identify the test statistic. (Round to two decimal places as needed.)

3. Use technology to identify the P-value. (Round to three decimal places as needed.)

4. What is the conclusion for this hypothesis test?

A. Fail to reject H 0. There is insufficient evidence to support the claim that the treatment group has errors that vary significantly more than the errors of the placebo group.
B. Reject H 0. There is insufficient evidence to support the claim that the treatment group has errors that vary significantly more than the errors of the placebo group.
C. Fail to reject H 0. There is sufficient evidence to support the claim that the treatment group has errors that vary significantly more than the errors of the placebo group.
D. Reject H 0. There is sufficient evidence to support the claim that the treatment group has errors that vary significantly more than the errors of the placebo group.

C.

D.

Identify the test statistic=______ (Round to two decimal places)

The P-value is=_____ ( round to three decimal places)

What is the concluion for this hypothesis test?

A. Fail to reject Upper H0. There is sufficient evidence to warrant rejection of the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150.

B. Reject Upper H 0. There is insufficient evidence to warrant rejection of the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150.

C. Fail to reject Upper H 0. There is insufficient evidence to warrant rejection of the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150.

D. Reject Upper H 0. There is sufficient evidence to warrant rejection of the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150.

P.S Can you please answer part A-D. I am very sick and I can't think right now. Please help me im begging you. My assignment is due tonight and Ive been stuck in bed all week. Please help. May god bless your soul! Please look at all pictures so you won't miss a part. Thank you. I'll appreciate your help.

Answer 1

Solution: Part-A: Men: n = 185 x1 = 15668.5 Si = 8632.7 Women: n2 = 211 x2 = 16215.9 S2 = 7301.1 Null Hypothesis: H:01 = ož Alternative Hypothesis: HQ:01 > ož Degrees of freedom for numerator = d fi = 185 - 1 = 184 Degrees of freedom for denominator = df2 = 211 – 1 = 210 Level of significance = a = 0.05 Test statistic: s? 8632.72 CV 7301.12 = 1.40 P-value: From F-distribution table; area under the curve to the right of (F = 1.40) and for degrees of freedom (184, 210), is given by: 0.0092 Hence, p-value = 0.0092 Conclusion: Since p-value <a, we reject the null hypothesis. Reject Ho. There is sufficient evidence to support the claim that the number of words spoken in a day by men vary more than the number of words spoken in a day by women.

Part-B: Ethanol: ni = 24 Si = 2.30 Placebo: n2 = 24 S2 = 0.85 Null Hypothesis: H.:01 = ož Alternative Hypothesis: Hą:07 > 02 Degrees of freedom for numerator = df1 = 24 – 1 = 23 Degrees of freedom for denominator = df2 = 24 – 1 = 23 Level of significance = a = 0.05 Test statistic: s? 2.302 F = = s 0.852 = 7.32 P-value: Using Excel's formula =F.DIST.RT(7.32, 23, 23), area in the right tail is given by: 0.0000 Hence, p-value = 0.0000 Conclusion: Since p-value <a, we reject the null hypothesis. Reject Ho. There is sufficient evidence to support the claim that the treatment group has errors that vary significantly more than the errors of the placebo group.

Part-C: Sham: ni = 20 X1 = 0.43 S1 = 1.31 Magnet: n2 = 20 X = 0.46 S2 = 0.94 Null Hypothesis: H.:01 = ož Alternative Hypothesis: Ha:01 > Degrees of freedom for numerator = dfı = 20 – 1 = 19 Degrees of freedom for denominator = dfz = 20 - 1 = 19 Level of significance = a = 0.05 Test statistic: s 1.312 F = 1 = 6022 = 1.94 1.94 0.942 P-value: From F-distribution table; area under the curve to the right of (F = 1.94) and for degrees of freedom (19, 19), is given by: 0.0788 Hence, p-value = 0.0788 Conclusion: Since p-value <a, we reject the null hypothesis. Reject Ho. There is sufficient evidence to support the claim that those given a sham treatment have pain reductions that vary more than those treated with magnets.

Part-D: The standard deviation for A.D. 150 sample is higher. Hence, we will treat it as first population and the 4000 B.C. as second population A.D. 150: ni = 30 X1 = 136.02 Si = 5.37 4000 B.C.: n2 = 30 x2 = 131.26 S2 = 5.11 Null Hypothesis: H.:01 = ož Alternative Hypothesis: Ha: 01 + 02 Degrees of freedom for numerator = df1 = 30 – 1 = 29 Degrees of freedom for denominator = df2 = 30 – 1 = 29 Level of significance = a = 0.05 Test statistic:

s? 5.372 5.112 = 1.10 P-value: From F-distribution table; area under the curve to the right of (F = 1.10) and for degrees of freedom (29, 29), is given by: 0.3996 The test is a two-tailed test. Hence, p-value = 0.3996 x 2 = 0.7992 Conclusion: Since p-value >a, we fail to reject the null hypothesis. Fail to reject Ho. There is insufficient evidence to warrant rejection of the claim that the variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150.

Answer 2

-2.88 ( round to two decimal places as needed)