Questions: Cost of a Movie Ticket The average cost of a movie ticket is 13. A random sample in a specific city of 41 movie theaters found that the average cost of a movie ticket was 13.12. If σ=0.38 and α=0.05, can it be concluded that the average cost of movie tickets in that city is not 13?
Transcript text: Cost of a Movie Ticket The average cost of a movie ticket is $\$ 13$. A random sample in a specific city of 41 movie theaters found that the average cost of a movie ticket was $\$ 13.12$. If $\sigma=\$ 0.38$ and $\alpha=0.05$, can it be concluded that the average cost of movie tickets in that city is not $\$ 13$ ?
Part: $0 / 5$
Solution
Solution Steps
Step 1: Define the Hypotheses
We want to test whether the average cost of movie tickets in the city is significantly different from \$13. The hypotheses are defined as follows:
Null Hypothesis (\(H_0\)): \(\mu = 13\)
Alternative Hypothesis (\(H_a\)): \(\mu \neq 13\)
Step 2: Calculate the Standard Error
The standard error (\(SE\)) is calculated using the formula:
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{0.38}{\sqrt{41}} \approx 0.0593
\]
Step 3: Calculate the Test Statistic
The Z-test statistic is calculated using the formula:
For a two-tailed test, the P-value is calculated as:
\[
P = 2 \times (1 - T(|z|)) \approx 0.0432
\]
Step 5: Compare the P-value with the Significance Level
The significance level (\(\alpha\)) is set at 0.05. We compare the P-value with \(\alpha\):
If \(P < \alpha\), we reject the null hypothesis.
If \(P \geq \alpha\), we fail to reject the null hypothesis.
In this case:
\[
0.0432 < 0.05
\]
Step 6: Conclusion
Since the P-value is less than the significance level, we reject the null hypothesis. This indicates that there is sufficient evidence to conclude that the average cost of movie tickets in the city is significantly different from \$13.
Final Answer
\(\boxed{\text{Reject the null hypothesis: The average cost of movie tickets in the city is significantly different from } \$13.}\)