STATISTICAL TECHNIQUES FOR BUSINESS AND
STATISTICAL TECHNIQUES FOR BUSINESS AND
17th Edition
ISBN: 9781307261158
Author: Lind
Publisher: MCG/CREATE
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Chapter 11, Problem 21CE
To determine

Check whether there is a difference in the mean number of times men and women order take-out dinners in a month.

Find the p-value.

Expert Solution & Answer
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Answer to Problem 21CE

The conclusion is that there is no difference in the mean number of times men and women order take-out dinners in a month.

The p-value is 0.063.

Explanation of Solution

Calculation:

The null and alternative hypotheses are stated below:

Let μ1 is the average men who live alone buy take-out dinner in a month and μ2 the average women who live alone buy take-out dinner in a month.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

H1:μ1μ2

Significance level, α:

It is given that the significance level, α=0.01.

Degrees of freedom:

The degrees of freedom is as follows:

df=n1+n22=35+402=73

Thus, the number of degrees of freedom is 73.

Step-by-step procedure to obtain the critical value using MINITAB software:

  • Choose Graph > Probability Distribution Plot choose View Probability > OK.
  • From Distribution, choose ‘t’ distribution.
  • In Degrees of freedom, enter 73.
  • Click the Shaded Area tab.
  • Choose P Value and both Tails for the region of the curve to shade.
  • Enter the probability value as 0.01.
  • Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 21CE , additional homework tip  1

From the MINITAB output, the critical value is ±2.645.

The decision rule is as follows:

If t <2.645, then reject the null hypothesis H0.

If t>2.645, then reject the null hypothesis H0.

Pooled estimate:

The pooled estimate of the population variance is as follows:

=(n11)s12+(n21)s22n1+n22

Substitute s1 as 4.48, s2 as 3.86, n1 as 35, and n2 as 40 in the formula given above as follows:

sp2=(351)(4.48)2+(401)(3.86)235+402=(34)(20.0704)+(39)(14.8996)25=1,263.47873=17.31

Test statistic:

The test statistic for the hypothesis test of μ1μ2 when σ1 and σ2 are unknown is as follows:

t=(x¯1x¯2)sp2(1n1+1n2)

Substitute x¯1 as 20.0704, x¯2 as 14.8996, sp2 as 17.31, n1 as 35, and n2 as 40 in the formula given above as follows:

t=24.5122.6917.31(135+140)=1.820.9273=1.820.9630=1.890

Thus, the test statistic is 1.890.

Decision:

The critical value is 2.645, and the value of the test statistic is 1.890.

The value of the test statistic is less than the critical value.

That is, Test statistic(1.890)<Criticalvalue(2.645).

From the decision rule, fail to reject the null hypothesis.

Conclusion:

Therefore, there is no evidence that the mean number of times men and women order take-out dinners in a month.

p-value:

Step-by-step procedure to obtain the p-value using MINITAB software:

  • Choose Graph > Probability Distribution Plot choose View Probability > OK.
  • From Distribution, choose ‘t’ distribution.
  • In Degrees of freedom, enter 73.
  • Click the Shaded Area tab.
  • Choose X Value and Two Tail for the region of the curve to shade.
  • Enter the X value as 1.890.
  • Click OK.

Output obtained using MINITAB software is given below:

STATISTICAL TECHNIQUES FOR BUSINESS AND, Chapter 11, Problem 21CE , additional homework tip  2

From the MINITAB output, the p-value is 0.03136.

p-value = 2(0.03136)=0.063

Thus, the p-value is 0.063.

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