Questions: FAA regulations define an airplane to be "on-time" if it takes off within 15 minutes of scheduled departure time. Happy Airlines advertises that at least 90% of its flights are on time. In a study conducted to test Happy Airlines' advertising claim, 50 of 56 randomly selected flights took off "on time". Assuming that the distribution is normal, what can be concluded at the 0.05 level of significance? H0: p >= 0.90 Ha: p Test statistic: = (t or z, then a number) Round your answer to 2 decimal places. p -value = Round your answer to four decimal places. p -value Interpretation: If the null hypothesis is , then the probability that the sample proportion is is . Round numbers to 4 decimals.

FAA regulations define an airplane to be "on-time" if it takes off within 15 minutes of scheduled departure time. Happy Airlines advertises that at least 90% of its flights are on time. In a study conducted to test Happy Airlines' advertising claim, 50 of 56 randomly selected flights took off "on time".

Assuming that the distribution is normal, what can be concluded at the 0.05 level of significance?

H0: p >= 0.90
Ha: p

Test statistic: = (t or z, then a number) Round your answer to 2 decimal places.
p -value = Round your answer to four decimal places.
p -value Interpretation: If the null hypothesis is ,
then the probability that the sample proportion is
is . Round numbers to 4 decimals.

Transcript text: FAA regulations define an airplane to be "on-time" if it takes off within 15 minutes of scheduled departure time. Happy Airlines advertises that at least $90 \%$ of its flights are on time. In a study conducted to test Happy Airlines' advertising claim, 50 of 56 randomly selected flights took off "on time". Assuming that the distribution is normal, what can be concluded at the 0.05 level of significance? \[ \begin{array}{l} H_{0}: p \geq 0.90 \\ H_{a}: p \end{array} \] Test statistic: $\square$ $=$ $\square$ (t or z, then a number) Round your answer to 2 decimal places. p -value $=$ $\square$ Round your answer to four decimal places. p -value Interpretation: If the null hypothesis is $\square$ , then the probability that the sample proportion is $\square$ $\square$ is $\square$ . Round numbers to 4 decimals.

Solution

Solution Steps

Step 1: Hypothesis Formulation

We are testing the claim made by Happy Airlines that at least $90\%$ of its flights are on time. The hypotheses are defined as follows:

Null Hypothesis ($H_0$): $p \geq 0.90$
Alternative Hypothesis ($H_a$): $p < 0.90$

Step 2: Sample Proportion Calculation

From the study, we have:

Number of flights that took off on time: $50$
Total number of flights sampled: $56$

The sample proportion ($\hat{p}$) is calculated as: \[ \hat{p} = \frac{50}{56} \approx 0.8929 \]

Step 3: Test Statistic Calculation

The test statistic for the proportion is calculated using the formula: \[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] Substituting the values:

$p_0 = 0.90$
$n = 56$

Calculating the standard error: \[ \text{Standard Error} = \sqrt{\frac{0.90(1 - 0.90)}{56}} \approx 0.1265 \]

Now substituting into the Z formula: \[ Z = \frac{0.8929 - 0.90}{0.1265} \approx -0.1782 \]

Step 4: P-value Calculation

The p-value associated with the test statistic $Z = -0.1782$ is calculated to be: \[ \text{P-value} \approx 0.4293 \]

Step 5: Critical Region and Decision

For a significance level of $\alpha = 0.05$ in a left-tailed test, the critical value is: \[ Z < -1.6449 \]

Since the calculated test statistic $Z = -0.1782$ does not fall into the critical region, we fail to reject the null hypothesis.

Step 6: Interpretation of Results

The interpretation of the p-value is as follows: If the null hypothesis is "fail to reject," then the probability that the sample proportion is $0.8929$ or more extreme is $0.4293$.