Answer:
So the p value obtained was a very low value and using the significance level given we have
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
estimated proportion of women indicated that they would be willing to give up some personal in order to make more money
is the value that we want to test
represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
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:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statistic, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
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 . The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
So the p value obtained was a very low value and using the significance level given we have
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 and
. And the critical value would be given by:
The confidence interval for the mean is given by the following formula:
If we replace the values obtained we got:
The 95% confidence interval would be given by (0.509;0.571)