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 time in order to make more money. The sample was selected in a way that was designed to produce a sample that was representative of women in the targeted age group. (a) Do the sample data provide convincing evidence that the majority of women 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 α = 0.01. (Round your test statistic to two decimal places and your P-value to four decimal places.)

Answer 1

Answer:

$z=\frac{0.540-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53$  

$p_v =P(Z>2.53)=0.0057$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the true proportion is not significantly higher than 0.5.  

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

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

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

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

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the majority of women 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 $p_o$.

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =100*0.5=500>10$

$n(1-p_o)=1000*(1-0.5)=500>10$

3) Calculate the statistic  

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

$z=\frac{0.540-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53$  

4) 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 provided $\alpha=0.01$. The next step would be calculate the p value for this test.  

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

$p_v =P(Z>2.53)=0.0057$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the true proportion is not significantly higher than 0.5.