At 5% significance level, larger proportion of military personnel students protect their profiles on social-networking sites than young adults with profiles on social-networking sites.

Step-by-step explanation:

let p be the proportion of military personnel students who restrict access to their profiles. Then null and alternative hypotheses are:

$H_{0}$ : p=0.67 (67%)

$H_{a}$ : p>0.67

We need to calculate z-statistic of sample proportion:

z= $\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

p(s) is the sample proportion of military students who restrict access to their profiles ( $\frac{78}{100}$ =0.78)

p is the proportion assumed under null hypothesis. (0.67)

N is the sample size (100)

Then z= $\frac{0.78-0.67}{\sqrt{\frac{0.67*0.33}{100} } }$ ≈ 2.34

The corresponding p-value is 0.0096. Since 0.0096<0.05 (significance level) we can reject the null hypothesis and conclude at 5% significance level that larger proportion of military personnel students protect their profiles on social-networking sites than young adults with profiles on social-networking sites.

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3 years ago