Question

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

0.050.05

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.

	n	x overbarx	s
Men	185185	15 comma 668.715,668.7	8 comma 632.88,632.8
Women	211211	16 comma 215.716,215.7	7 comma 301.77,301.7

What are the null and alternative​ hypotheses?

A.

Upper H 0H0​:

sigma Subscript 1 Superscript 2σ21not equals≠sigma Subscript 2 Superscript 2σ22

Upper H 1H1​:

sigma Subscript 1 Superscript 2σ21equals=sigma Subscript 2 Superscript 2σ22

B.

Upper H 0H0​:

sigma Subscript 1 Superscript 2σ21equals=sigma Subscript 2 Superscript 2σ22

Upper H 1H1​:

sigma Subscript 1 Superscript 2σ21greater than>sigma Subscript 2 Superscript 2σ22

C.

Upper H 0H0​:

sigma Subscript 1 Superscript 2σ21equals=sigma Subscript 2 Superscript 2σ22

Upper H 1H1​:

sigma Subscript 1 Superscript 2σ21less than<sigma Subscript 2 Superscript 2σ22

D.

Upper H 0H0​:

sigma Subscript 1 Superscript 2σ21equals=sigma Subscript 2 Superscript 2σ22

Upper H 1H1​:

sigma Subscript 1 Superscript 2σ21not equals≠sigma Subscript 2 Superscript 2σ22

Identify the test statistic.

Fequals=. 97.97

​(Round to two decimal places as​ needed.)

Use technology to identify the​ P-value.

The​ P-value is

nothing.

​(Round to three decimal places as​ needed.)

Answer 1

Null and alternative hypothesis:  
Hₒ : σ₁ = σ₂  
H₁ : σ₁ > σ₂  

Answer B)

Test statistic:  
F = s₁² / s₂² = 8632.8² / 7301.7² =    1.40
Degree of freedom:  
df₁ = n₁-1 =    184
df₂ = n₂-1 =    210

  
P-value :  
P-value = F.DIST.RT(1.3978, 184, 210) =    0.009
Conclusion:  
As p-value < α, we reject the null hypothesis

THANKS

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