Question

On a research it was found that 47% of residents travelers were women and 53% men. A random sample of 500 travelers on a large airline revealed that of those 500, 263 were women. Using a=0.05 compute CV, and TV, and state the hypotheses and your conclusion.

Answer 1

Probability of women residents $p=0.47$

Also, 263 were women from sample of 500 , $\hat{p}=\dfrac{263}{500}=0.526$

Let $H_o:p=\hat{p}$ be the Null hypothesis.

$H_a:p\neq \hat{p}$ be the alternate hypothesis.

$\alpha=0.05,$

$z_{\frac{\alpha}{2}}=z_{0.025}=1.96$

Critical value is 1.96.

Calculating Test value(TV)-

$z=\dfrac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$=\dfrac{0.526-0.47}{\sqrt{\frac{0.47(1-0.47)}{500}}}=\dfrac{0.056}{0.02232}=2.5089$

Conclusion: As the test value is greater than the critical value, So Null Hypotesis is rejected i.e $p\neq \hat{p}$