Question

More people are using social media to network, rather than phone calls or e-mails (US News & World Report, October 20, 2010). From an employment perspective, jobseekers are no longer calling up friends for help with job placement, as they can now get help online. In a recent survey of 150 jobseekers, 67 said they used LinkedIn to search for jobs. A similar survey of 140 jobseekers, conducted three years ago, had found that 58 jobseekers had used LinkedIn for their job search. Use Table 1.
Let p1 represent the population proportion of recent jobseekers and p2 the population proportion of job seekers three years ago. Let recent survey and earlier survey represent population 1 and population 2, respectively.a. Set up the hypotheses to test whether there is sufficient evidence to suggest that more people are now using LinkedIn to search for jobs as compared to three years ago.
A. H0: p1 − p2 ≥ 0; HA: p1 − p2 < 0
B. H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0
C. H0: p1 − p2 = 0; HA: p1 − p2 ≠ 0
b. Calculate the value of the test statistic. (Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)c. Calculate the critical value at the 5% level of significance. (Round your answer to 3 decimal places.)d. Interpret the results.
A. Do not reject H0; there is no increase in the proportion of people using LinkedIn
B. Do not reject H0; there is an increase in the proportion of people using LinkedIn
C. Reject H0; there is no increase in the proportion of people using LinkedIn
D. Reject H0; there is an increase in the proportion of people using LinkedIn

Answer 1

Answer:

B. H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0

$z=\frac{0.447-0.414}{\sqrt{0.431(1-0.431)(\frac{1}{150}+\frac{1}{140})}}=0.57$    

$z_{crit}=1.64$

A. Do not reject H0; there is no increase in the proportion of people using LinkedIn

Step-by-step explanation:

1) Data given and notation  

$X_{1}=67$ represent the number of recent jobseekers

$X_{2}=58$ represent the number of job seekers three years ago.

$n_{1}=150$ sample of recent jobseekers selected  

$n_{2}=140$ sample of job seekers three years ago selected  

$p_{1}=\frac{67}{150}=0.4468$ represent the proportion of recent jobseekers

$p_{2}=\frac{58}{140}=0.4143$ represent the proportion of job seekers three years ago

z would represent the statistic (variable of interest)  

$p_v$ represent the value for the test (variable of interest)  

$\alpha=0.05$ significance level given

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if "More people are using social media to network, rather than phone calls or e-mails", the system of hypothesis would be:  

Null hypothesis:$p_{1} - p_{2} \leq 0$  

Alternative hypothesis:$p_{1} - p_{2} > 0$  

We need to apply a z test to compare proportions, and the statistic is given by:  

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$   (1)  

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{67+58}{150+140}=0.4310$  

3) Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

$z=\frac{0.4468-0.4143}{\sqrt{0.4310(1-0.4310)(\frac{1}{150}+\frac{1}{140})}}=0.5671$    

In order to find the critical value since we have a right tailed test the we need to find a value on the z distribution that accumulates 0.05 of the area on the right tail, and this value is$z_{crit}=1.64$.

4) Statistical decision

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

$p_v =P(Z>0.5671)= 0.285$  

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

So the correct conclusion would be:

A. Do not reject H0; there is no increase in the proportion of people using LinkedIn