Question

Hard times In June 2010, a random poll of 800 working men found that 9% had taken on a second job to help pay the bills. (www.careerbuilder) a) Estimate the true percentage of men that are taking on second jobs by constructing a 95% confidence interval. b) A pundit on a TV news show claimed that only 6% of work-ing men had a second job. Use your confidence interval to test whether his claim is plausible given the poll data.

Answer 1

Answer:

a) The 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

b) Since the confidence interval NOT contains the value 0.06 we  have anough evidence to reject the claim at 5% of significance.  

Step-by-step explanation:

Previous concepts

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".  

$p$ represent the real population proportion of interest  

$\hat p =0.09$ represent the estimated proportion for the sample  

n=800 is the sample size required  

$z$ represent the critical value for the margin of error  

Confidence =0.95 or 95%

Part a

The population proportion have the following distribution  

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

The confidence interval would be given by this formula  

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

The margin of error is given by :

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

$Me=1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0198$

And replacing into the confidence interval formula we got:  

$0.09 - 1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0702$  

$0.09 + 1.96 \sqrt{\frac{0.108(1-0.09)}{800}}=0.1098$  

And the 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

Part b

Since the confidence interval NOT contains the value 0.06 we  have anough evidence to reject the claim at 5% of significance.