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 308 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.
1) State the null and alternative hypothesis.
2) Find the value of the test statistic.
3) Find the p-value.
B) Conduct a hypothesis test to determine if the results of the survey justify concluding that more than 70% of 22- to 30-year-old job seekers have electronically submitted a resume to an employer. Using α = 0.05, what is your conclusion?
1) State the null and alternative hypothesis.
2) Find the value of the test statistic.
3) Find the p-value.

A) i) Null hypothesis; p ≤ 0.79

Alternative hypothesis; p > 0.79

ii) test statistic = 2

iii) P-value = 0.02275

B) i) Null hypothesis; p ≤ 0.70

Alternative hypothesis; p > 0.70

ii) test statistic = 2.5

iii) P-value = 0.00621

Step-by-step explanation:

A) We are given;

Population proportion; p = 79% = 0.79

308 of the 370 respondents use social media in their job search. Thus;

Sample proportion; p^ = 308/370 = 0.8324

Let's state the hypotheses;

Null hypothesis; p ≤ 0.79

Alternative hypothesis; p > 0.79

Formula for the test statistic is;

z = (p^ - p)/√(p(1 - p)/n)

z = (0.8324 - 0.79)/√(0.79(1 - 0.79)/370)

z = 0.0424/0.021175

z = 2

From online p-value from z-score calculator attached and using; z = 2; significance value = 0.05; and one tailed distribution, we have;

P-value = 0.02275

B) similar to answer A above, Population proportion; p = 70% = 0.70

But 281 of the 370 respondents use social media in their job search. Thus;

Null hypothesis; p ≤ 0.70

Alternative hypothesis; p > 0.70

Also; Sample proportion; p^ = 281/370 = 0.7595

Formula for the test statistic is;

z = (p^ - p)/√(p(1 - p)/n)

z = (0.7595 - 0.70)/√(0.70(1 - 0.70)/370)

z = 0.0595/0.023824

z = 2.5

From online p-value from z-score calculator attached and using; z = 2.5; significance value = 0.05; and one tailed distribution, we have;

P-value = 0.00621

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)