Question

13 multiple choice 1 point
conduct the appropriate hypothesis test and compute the test statistic.

sunshine air has determined that their no - show rate for passengers booked on a flight is 6%. the airline has recently increased the cost of its travel insurance and and suspects the no - show rate for passengers will decrease.

a random sample of 380 reservations resulted in 18 no - shows.

at the 0.10 significance level, is there strong enough sample evidence to suggest that the no - show rate in less than 6%?

○ no, because the test statistic $z^{*} = - 1.04$.
○ yes, because the test statistic $z^{*} = - 1.20$.
○ yes, because the test statistic $z^{*} = - 1.04$.
○ no, because the test statistic $z^{*} = - 1.20$.

Explanation:

Step1: Identify Hypotheses

Null: \( H_0: p = 0.06 \), Alternative: \( H_a: p < 0.06 \) (left - tailed test).

Step2: Calculate Sample Proportion

Sample proportion \( \hat{p}=\frac{x}{n}=\frac{18}{380}\approx0.0474 \).

Step3: Compute Test Statistic

The formula for the z - test statistic for a proportion is \( z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \)
Substitute \( \hat{p} = 0.0474 \), \( p_0=0.06 \), \( n = 380 \)
First, calculate the denominator: \( \sqrt{\frac{0.06\times(1 - 0.06)}{380}}=\sqrt{\frac{0.06\times0.94}{380}}=\sqrt{\frac{0.0564}{380}}\approx\sqrt{0.0001484}\approx0.0122 \)
Then, calculate the numerator: \( 0.0474-0.06=- 0.0126 \)
Now, find the z - statistic: \( z=\frac{-0.0126}{0.0122}\approx - 1.04 \)

Step4: Determine Significance and Decision

Significance level \( \alpha = 0.10 \). For a left - tailed test, the critical value \( z_{\alpha}=-1.28 \) (from standard normal table).
Our test statistic \( z=-1.04 \) is greater than \( - 1.28 \) (i.e., \( - 1.04>-1.28 \)), so we fail to reject the null hypothesis. So the answer is "No, because the test statistic \( z^{*}=-1.04 \)".

Answer:

No, because the test statistic \( z^{*}=-1.04 \) (the first option: No, because the test statistic \( z^{*}=-1.04 \))