Question

of u.s. residents traveling overseas, 47% were women and 53% were men. a random sample of 500 travelers on a large airline revealed that of those 500, 263 were women. does this differ from the national percentage at the 0.05 level of significance? state the type of test. (example: one tail to the right/one tail to the left/ two tails) two tails state the critical value. (example: z=2.58 or +/- - 2.58) z = 1.96 state the test statistic. (example: z=2.76 or - 2.76) z = 2.51 state the decision. (example: reject/fail to reject) reject

Explanation:

Step1: Calculate sample proportion

The sample proportion \( \hat{p}=\frac{263}{500} = 0.526\)

Step2: Identify population proportion

The population proportion \(p = 0.47\)

Step3: Calculate standard error

The standard error \(SE=\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.47\times(1 - 0.47)}{500}}=\sqrt{\frac{0.47\times0.53}{500}}\approx\sqrt{\frac{0.2491}{500}}\approx\sqrt{0.0004982}\approx0.0223\)

Step4: Calculate z - statistic

The z - statistic \(z=\frac{\hat{p}-p}{SE}=\frac{0.526 - 0.47}{0.0223}=\frac{0.056}{0.0223}\approx2.51\)

Step5: Determine critical value

For a two - tailed test with \(\alpha = 0.05\), the critical values are \(z=\pm1.96\)

Step6: Make a decision

Since \(|z| = 2.51>1.96\), we reject the null hypothesis.

Answer:

• Type of test: two tails
• Critical value: \(z = \pm1.96\)
• Test statistic: \(z = 2.51\)
• Decision: reject