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algol [13]
3 years ago
7

The Pew Internet and American Life Project reported Wednesday, April 18th that two- thirds (67%) of young adults with profiles o

n social-networking sites have restricted access to their profiles by requiring passwords or making them available only to friends on an approved list ("Study: Teens protecting their profiles" POSTED: 10:54 a.m. EDT, April 19, 2007 on CNN.com). A researcher believes that military personnel are more careful with keeping themselves safe online and surveys a random sample of 100 enlisted military personnel students and finds that 78 students restrict access to their profiles. Does this indicate at the 5% significance level that a larger proportion of military personnel protect their profiles on social-networking sites?
Mathematics
1 answer:
dlinn [17]3 years ago
3 0

Answer:

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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Answer:

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Step-by-step explanation:

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