Question

In a survey of 1000 women age 22 to 35 who work full time, 540 indicated that they would be willing to give up some personal in order to make more money. The sample was selected in a way that was designed to produce a sample that was representative of woman in the target age group. Does the sample data provide convincing evidence that the majority of woman age 22 to 35 who work full-time would be willing to give up some personal time for more money? Test the relevant hypotheses using .
(You must check the conditions for normality of the sampling distribution, state the hypotheses, test statistic and P-value.) Next construct a 95% confidence interval based on this data.

Answer 1

Answer:

$z=\frac{0.54 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.530$  

$p_v =P(Z>2.530)=0.0057$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of women indicated that they would be willing to give up some personal in order to make more money  is significantly higher than 0.5.  

The 95% confidence interval would be given by (0.509;0.571)

Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

X=540 represent the women indicated that they would be willing to give up some personal in order to make more money

$\hat p=\frac{540}{1000}=0.54$ estimated proportion of women indicated that they would be willing to give up some personal in order to make more money

$p_o=0.5$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the majority of woman age 22 to 35 who work full-time would be willing to give up some personal time for more money.:  

Null hypothesis:$p \leq 0.5$  

Alternative hypothesis:$p > 0.5$  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.54 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.530$  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

$p_v =P(Z>2.530)=0.0057$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of women indicated that they would be willing to give up some personal in order to make more money  is significantly higher than 0.5.  

Confidence interval

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 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

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:

$0.54 - 1.96\sqrt{\frac{0.54(1-0.54)}{1000}}=0.509$

$0.54 + 1.96\sqrt{\frac{0.54(1-0.54)}{1000}}=0.571$

The 95% confidence interval would be given by (0.509;0.571)