Question

A.

4 9.4.5-T Question Help Researchers measured skulls from different time periods in an attempt to determmine whether interbreading of cultures occurred. Results are given below. 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 variation of maximal skull breadths in 4000 B.C. is the same as the variation in A.D. 150. п s х 4000 B.C A.D. 150 30 131.04 mm 5.14 mm 136.01 mm 5.39 mm 30 What are the null and altemative hypotheses? OA Ho: ов. н; с? - о? H: D. Ho: 0 H:0 03 ос. На о? - о H: 020 Identify the test statistic. F (Round to two decimal places as needed.) Enter your answer in the answer box and then click Check Answer. parts remaining Clear All Check Answer

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.

B.

C.

D.

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.

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

P-Value=_____ (Round to four decimal places)

What is the conclusion for his test?

Does it reject or fail to reject and is/is not sufficient evidence?

Answer 1

Here we have п, %330, S, 35.14 п, %330, S,-5.39 Hypotheses are: Level of significance: a =0.05 Test is two tailed. Degrees of freedom: df1=n1-1=29, df2=n2-1=29 Test Statistics: 2 F = (5.14^2)/(5.39^2) =0.909 Critical value(CVs): 0.48, 2.1 Rejection Region: If F< 0.48 or F > 2.1, Reject HO Decision: Since test statistics does not lie in rejection region so we fail to reject the null hypothesis P-value: P-value-0.799 Decision: Since p-value is not less than level of significance so we fail to reject the null hypothesis. =FINV((1-(0.05/2)),29,29) and -FINV((0.05/2),29,29) Excel function for critical value: 2*MIN(1-FDIST(0.909,29,29)), FDIST(0.909,29,29)) Excel function for p-value:

Test statistics: F = 0.91

The p-value is =0.799

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.

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B:

Here we have 22,s 2.2 22, S2 0.85 п, п, Hypotheses are: Level of significance: a 0.05 Test is right tailed. Degrees of freedom: df1=n1-1-21, df2=n2-1=21 Test Statistics: (2.2 2)/(0.85A2) $2 F = 2 =6.7 Critical value(CVs): 2.08 Rejection Region: If F> 2.08, Reject HO Decision: Since test statistics lies in rejection region so we the null hypothesis P-value: P-value 0 Decision: Since p-value is less than level of significance so we reject the null hypothesis. Excel function for critical value: =FINV(0.05,21,21) =FDIST (6.7,21,21) Excel function for p-value:

Correct option is B.