Answer to Question #306006 in Statistics and Probability for [e-mail]

Question

A report says that 80% of India’s females aged 15-59 are not currently engaged in the workforce. A national agency has an opinion that this percentage may be even more. To validate its opinion, the agency did a survey of randomly chosen 1200 females of the age group 15-59 from the different parts of the country and found 228 females working. Do the figures of the survey help the agency in validation of its opinion?

Answer 1

Hypotheses

H₀: p=0.8

H_a: p>0.8

α = 0.05

Test statistic

We first check for the sample size to determine if the Z statistic can be used. In this case, the condition that the min(n*p, n*p(1-p))≥5 must be met for the Z statistic to be used.

Given n=1200 and p=0.8

min(1200*0.8, 1200*0.8*0.2) = min(960, 192) = 192

Since the condition is met, the appropriate test statistic is:

z = (p̂ - p) /$\sqrt{p*(1-p)/n}$ where:

p̂ = sample proportion, p = hypothesized population proportion

n = sample size

Decision Rule

This is a upper tailed test. By using the the Z statistic and 5% level of significance level, we find that Z = 1.645 from the Z reference table. Therefore, the decision rule is to reject the null hypothesis if the computed test statistic Z ≥1.645

Computed test statistic

Let x be the number of females in the age group 15-59 who are not working

Given that 228 females out of 1200 are found not working, then

Then x=1200-228 = 972

Also given that n=1200

Then,

p̂ = x/n = 972/1200 = 0.81

and p=0.8

Therefore

z = (p̂ - p) /$\sqrt{p(1-p)/n}$ = (0.81 - 0.8) /$\sqrt{(0.8*0.2)/1200}$

= 0.8660

Decision

Since the computed test statistic Z= 0.8660<1.645, we fail to reject the null hypothesis.

Conclusion

At 5% level of significance, the figures from opinion do not provide sufficient evidence to conclude that more than 80% of India’s females aged 15-59 are not currently engaged in the workforce.