Question

Labor statistics indicate that 77% of all U.S. cashiers and servers are women. A random sample of cashiers and servers in a particular metropolitan area found that 112 of 150 cashiers and 150 of 200 servers were women. At the 0.05 level of significance, is there sufficient evidence to conclude that a difference exists between this metropolitan area and the larger U.S. population? What is the test statistic? Round to 2 decimal places.

Answer 1

The test statistic for the difference in proportions between the metropolitan area and the U.S. population is approximately ( -0.11 ). Since this value does not exceed the critical value of
`\( \pm 1.96 \)` at the 0.05 significance level, there is insufficient evidence to conclude a significant difference. Therefore, we fail to reject the null hypothesis.

Explanation:

To determine whether there is sufficient evidence to conclude that a difference exists between this metropolitan area and the larger U.S. population, we can perform a hypothesis test for the difference in proportions.

Let:

`\( p_1 \)` be the proportion of women in the metropolitan area.

`\( p_2 \)` be the proportion of women in the larger U.S. population.

The null hypothesis
`(\( H_0 \))` is that there is no difference between the proportions, and the alternative hypothesis
`(\( H_1 \))` is that there is a significant difference.

The formula for the test statistic for the difference in proportions ( z ) is given by:

`\[ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{p(1-p)\left((1)/(n_1) + (1)/(n_2)\right)}} \]`

Where:

`\( \hat{p}_1 \)` and
`\( \hat{p}_2 \)` are the sample proportions of women in the metropolitan area and the U.S. population, respectively.

( p ) is the combined sample proportion.

`\( n_1 \)` and
`\( n_2 \)` are the sample sizes for the metropolitan area and the U.S. population, respectively.

First, let's calculate
`\( \hat{p}_1 \)`,
`\( \hat{p}_2 \)`, ( p ), and then plug them into the formula to find the test statistic.

`\[ \hat{p}_1 = (112)/(150) = 0.7467 \]`

`\[ \hat{p}_2 = (150)/(200) = 0.75 \]`

`\[ p = (112 + 150)/(150 + 200) = (262)/(350) = 0.7486 \]`

Now, we can calculate the test statistic:

`\[ z = \frac{(0.7467 - 0.75)}{\sqrt{0.7486(1-0.7486)\left((1)/(150) + (1)/(200)\right)}} \]`

Calculate the values and round the test statistic to two decimal places. If the absolute value of the test statistic is greater than the critical value for a two-tailed test at the 0.05 significance level, we reject the null hypothesis.

Note: The critical value for a two-tailed test at the 0.05 significance level is approximately
`\( \pm 1.96 \)`.

Answer 2

We conclude that there is no difference between this metropolitan area and the larger U.S. population.

Explanation:

We are given that Labor statistics indicate that 77% of all U.S. cashiers and servers are women.

A random sample of cashiers and servers in a particular metropolitan area found that 112 of 150 cashiers and 150 of 200 servers were women.

Let p = proportion of all cashiers and servers who are women in metropolitan area.

So, Null Hypothesis,
`H_0` : p = 77% {means that there is no difference between this metropolitan area and the larger U.S. population}

Alternate Hypothesis,
`H_A` : p
`\\eq` 77% {means that there is a difference between this metropolitan area and the larger U.S. population}

The test statistics that would be used here One-sample z proportion statistics;

T.S. =
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = sample proportion of all cashiers and servers who are women in metropolitan area =
`(112+150)/(150+200)` = 0.75

n = sample of cashiers and servers = 150 + 200 = 350

So, test statistics =
`\frac{0.75-0.77}{\sqrt{(0.75(1-0.75))/(350) } }`

= -0.86

The value of z test statistics is -0.86.

Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference between this metropolitan area and the larger U.S. population.