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nikdorinn [45]
3 years ago
5

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

lowing data have been gathered. Downtown Store North Mall Store Sample size 25 20 Sample mean $9 $8 Sample standard deviation $2 $1 A 95% interval estimate for the difference between the two population means is a. .071 to 1.929. b. 1.09 to 4.078. c. 1.078 to 2.922. d. .226 to 1.774.
Mathematics
1 answer:
Julli [10]3 years ago
4 0

Answer:

0.071,1.928

Step-by-step explanation:

                                                Downtown Store   North Mall Store

Sample size   n                             25                        20

Sample mean \bar{x}                         $9                        $8

Sample standard deviation  s       $2                        $1

n_1=25\\n_2=20

\bar{x_1}=9\\ \bar{x_2}=8

s_1=2\\s_2=1

x_1-x_2=9-8=1

Standard error of difference of means = \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}

Standard error of difference of means = \sqrt{\frac{2^2}{25}+\frac{1^2}{20}}

Standard error of difference of means = 0.458

Degree of freedom = \frac{\sqrt{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}

Degree of freedom = \frac{\sqrt{(\frac{2^2}{25}+\frac{1^2}{20}})^2}{\frac{(\frac{2^2}{25})^2}{25-1}+\frac{(\frac{1^2}{20})^2}{20-1}}

Degree of freedom =36

So, z value at 95% confidence interval and 36 degree of freedom = 2.0280

Confidence interval = (x_1-x_2)-z \times SE(x_1-x_2),(x_1-x_2)+z \times SE(x_1-x_2)

Confidence interval = 1-(2.0280)\times 0.458,1+(2.0280)\times 0.458

Confidence interval = 0.071,1.928

Hence Option A is true

Confidence interval is  0.071,1.928

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