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3 years ago

42 1/2 < or > 42.500 = Help me please

Mathematics

1 answer:

KIM [24]3 years ago

3 0

Answer:

42 1/2>42.500

Step-by-step explanation:

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3 years ago

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

liberstina [14]

Answer:

Step-by-step explanation:

Hello!

You have two populations of interest and want to compare them. If you define the study variables as:

X₁: average hourly wages of an employee of the Downtown store.

n₁= 25

X[bar]₁= $9

S₁= $2

X₂: average hourly wages of an employee of the North Mall store.

n₂= 20

X[bar]₂= $8

S₂= $1

Both samples taken are independent, assuming that both populations are normal and that their population variances are equal I'll use the Student's-t statistic with a pooled sample variance to calculate the Confidence interval:

95% CI for μ₁ - μ₂

(X[bar]₁-X[bar]₂) ± $t_{n_1+n_2-2; 1-\alpha /2} * (Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } )$

$Sa^2= \frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}$

$Sa^2= \frac{24*4+19*1}{25+20-2}= 2.67$

Sa= 1.64

$t_{n_1-n_2-2;1-\alpha /2} = t_{43; 0.975} = 2.017$

(9-8)±2.017* $(1.64*\sqrt{\frac{1}{25} +\frac{1}{20} } )$

[0.007636;1.9923]

I hope it helps!

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Answer:

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4 years ago

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