Cinemark Theaters knows that a certain hit movie ran an average of 84 days in each city, and the corresponding standard deviation was 10 days. The manager of the southeastern part of Dallas was interested in comparing the movie’s popularity in his region with that in all of Cinemark’s other theaters. He randomly chose 75 theaters in his region and found that they ran the movie an average of 81.5 days.
At a 1% significance level, test these hypotheses.
1% significance level means 0.01 not 0.1.
To test these hypotheses, we can use a one-sample t-test.
The null hypothesis (H0) is that there is no difference in the average number of days the movie ran in the southeastern part of Dallas compared to all of Cinemark's other theaters. The alternative hypothesis (H1) is that there is a difference.
Given:
Sample mean (x̄) = 81.5
Population mean (μ) = 84 (known)
Sample size (n) = 75
Population standard deviation (σ) = 10 (known)
The test statistic for a one-sample t-test is calculated using the formula:
t = (x̄ - μ) / (σ / √n)
Plugging in the values, we get:
t = (81.5 - 84) / (10 / √75)
Calculating this expression:
t ≈ -2.12132
To find the critical value at a 1% significance level, we need to look up the t-distribution table using the degrees of freedom, which is n - 1 = 75 - 1 = 74. From the table, we find that the critical value for a one-tailed test at a 1% significance level is approximately -2.63.
Since -2.12132 > -2.63, the test statistic does not fall in the critical region. Therefore, we fail to reject the null hypothesis.
Conclusion:
At a 1% significance level, there is not enough evidence to conclude that the average number of days the movie ran in the southeastern part of Dallas is significantly different from all of Cinemark's other theaters.
To test these hypotheses, we need to perform a hypothesis test using the sample mean and standard deviation provided.
Here are the null and alternative hypotheses:
Null hypothesis (H0): The mean running time in the southeastern part of Dallas is equal to the mean running time in all of Cinemark's other theaters.
Alternative hypothesis (Ha): The mean running time in the southeastern part of Dallas is different from the mean running time in all of Cinemark's other theaters.
To perform the hypothesis test, we will use the z-test since the sample size is large enough (n > 30) and we know the population standard deviation.
Here are the steps to perform the z-test:
Step 1: Define the significance level. In this case, the significance level is 1% or 0.01.
Step 2: Calculate the test statistic. The test statistic for a z-test is calculated using the formula:
z = (x - μ) / (σ / √n)
Where:
x = sample mean
μ = population mean
σ = population standard deviation
n = sample size
In this case:
x = 81.5 (sample mean)
μ = 84 (population mean)
σ = 10 (population standard deviation)
n = 75 (sample size)
Substituting the values into the formula, we get:
z = (81.5 - 84) / (10 / √75)
Step 3: Determine the critical value(s). Since the alternative hypothesis is two-tailed, we need to find the critical values for the given significance level. The critical values mark the rejection region(s) in the distribution.
To find the critical values, we can use a z-table or a calculator. For a 1% significance level divided equally between the two tails, we find the z-critical values for α/2 = 0.01/2 = 0.005.
Using the z-table or calculator, we find that the z-critical value for 0.005 on each tail is approximately ±2.58.
Step 4: Compare the test statistic with the critical value(s). If the test statistic falls within the rejection region(s), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, we compare the test statistic (z) with the critical value (-2.58 and 2.58). If z is outside the range of -2.58 to 2.58, we reject the null hypothesis. If z is within the range, we fail to reject the null hypothesis.
Step 5: Calculate the p-value. The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. We compare the p-value with the significance level to determine if the results are statistically significant.
For a two-tailed test, we need to calculate the p-value as the probability of obtaining a z-value as extreme as the one we calculated (in either tail). We can find this probability using a z-table or calculator.
Step 6: Make the decision. If the p-value is less than the significance level (0.01), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
That's how you test these hypotheses using a z-test. You can now proceed with the calculations and decision-making process.
You can use a one-sample z-test.
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
With your data:
z = (81.5 - 84)/(10/√75) = ?
I'll let you finish the calculation.
If you use a 1% significant level for a two-tailed test, then the cutoff or critical value from a z-table would be ±2.58. Does your test statistic calculated above exceed either the plus or minus critical value? Draw your conclusions.
I hope this will help.