Question

According to Inc, 79% of job seekers used social media in their job search in 2018. Many believe this number is inflated by the proportion of 22- to 30-year-old job seekers who use social media in their job search. Suppose a survey of 22- to 30-year-old job seekers showed that 314 of the 370 respondents use social media in their job search. In addition, 281 of the 370 respondents indicated they have electronically submitted a resume to an employer. (a) Conduct a hypothesis test to determine if the results of the survey justify concluding the proportion of 22- to 30-year-old job seekers who use social media in their job search exceeds the proportion of the population that use social media in their job search. Use α = 0.05. State the null and alternative hypothesis. (Enter != for ≠ as needed.)

Answer 1

Answer:

Null hypothesis:$p\leq 0.79$  

Alternative hypothesis:$p > 0.79$  

$z=\frac{0.849 -0.79}{\sqrt{\frac{0.79(1-0.79)}{370}}}=2.786$  

$p_v =P(z>2.786)=0.00267$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.79.  

Step-by-step explanation:

1) Data given and notation

n=750 represent the random sample taken

X=314 represent the respondents that use social media in their job search.

$\hat p=\frac{314}{370}=0.849$ estimated proportion of respondents that use social media in their job search

$p_o=0.79$ 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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is hgiher than 0.79.:  

Null hypothesis:$p\leq 0.79$  

Alternative hypothesis:$p > 0.79$  

When we conduct a proportion test we need to use the z statisticc, 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 $p_o$.

3) Calculate the statistic  

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

$z=\frac{0.849 -0.79}{\sqrt{\frac{0.79(1-0.79)}{370}}}=2.786$  

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

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

$p_v =P(z>2.786)=0.00267$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.79.