Question

The countries of Europe report that 48% of the labor force is female. The United Nations wonders if the percentage of females in the labor force is the same in the United States. Representatives from the United States Department of Labor plan to check a random sample of over 10,000 employment records on file to estimate a percentage of females in the United States labor force.
a) The representatives from the department of labor want to estimate the percentage of females in the US labor force with 90% confidence. In a sample of 525 employment records, they found that 229 of the workers were female. Create the 90% confidence interval. Show all work. Assume conditions are met.
b) Interpret the meaning of the confidence interval.
c) Should the representatives from the Department of Labor conclude that the percentage of females in their labor force is lower than Europe's rate of 48%?

Answer 1

Answer:

a) $0.436 - 1.64\sqrt{\frac{0.436(1-0.436)}{525}}=0.400$

$0.436 + 1.64\sqrt{\frac{0.436(1-0.436)}{525}}=0.472$

The 90% confidence interval would be given by (0.400;0.472)

b) We are confident at 90% that the true proportion of female in the Labor Force is between 0.400 and 0.472

c) Null hypothesis:$p\geq 0.48$  

Alternative hypothesis:$p < 0.48$  

For this case since the upper limit of the confidence interval is lower than 0.48 (0.472<0.48) we can conclude that the true proportion of female is actually lower than 0.48 or 48% at 10% of signficance.

Step-by-step explanation:

Part a

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$. And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The estimated proportion for female is $\hat p = \frac{229}{525}= 0.436$

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.436 - 1.64\sqrt{\frac{0.436(1-0.436)}{525}}=0.400$

$0.436 + 1.64\sqrt{\frac{0.436(1-0.436)}{525}}=0.472$

The 90% confidence interval would be given by (0.400;0.472)

Part b

We are confident at 90% that the true proportion of female in the Labor Force is between 0.400 and 0.472

Part c

We need to conduct a hypothesis in order to test the claim that the true proportion is lower than 0.48:  

Null hypothesis:$p\geq 0.48$  

Alternative hypothesis:$p < 0.48$  

For this case since the upper limit of the confidence interval is lower than 0.48 (0.472<0.48) we can conclude that the true proportion of female is actually lower than 0.48 or 48% at 10% of signficance.