How long does it take to have food delivered? A Chinese restaurant advertises that the mean delivery will be no more than 30` minutes. A random sample of delivery times is shown below. Based on this sample data, is the mean delivery time different than 30 minutes? Use a 5% level of significance. 21, 28, 22, 23, 30, 27, 29, 30, 32, 24, 32, 25, 26, 31.

To determine if the mean delivery time is different from 30 minutes, we can conduct a hypothesis test using the provided sample data. Here are the step-by-step instructions:

Step 1: State the null hypothesis (H0) and the alternative hypothesis (H1):
H0: The mean delivery time is 30 minutes.
H1: The mean delivery time is different from 30 minutes.

Step 2: Choose the appropriate statistical test.
Since we are comparing the mean of a single sample to a known value, we can use a one-sample t-test.

Step 3: Set the level of significance (α).
In this case, the significance level is given as 5%, which means α = 0.05.

Step 4: Identify the test statistic.
Since the population standard deviation is not known, we will use the t-distribution and calculate the t-statistic.

Step 5: Calculate the test statistic.
To calculate the t-statistic, we need the sample mean, sample standard deviation, number of observations, and the hypothesized population mean.

Sample mean (x̄) = (21 + 28 + 22 + 23 + 30 + 27 + 29 + 30 + 32 + 24 + 32 + 25 + 26 + 31) / 14 = 27.29
Sample standard deviation (s) = 3.76
Number of observations (n) = 14
Hypothesized population mean (μ0) = 30 (given)

t-statistic = (x̄ - μ0) / (s / √n)
t-statistic = (27.29 - 30) / (3.76 / √14)

Step 6: Determine the critical value.
Since the alternative hypothesis is two-sided (mean delivery time could be greater or less than 30 minutes), we will use a two-tailed t-test.
We need to find the critical t-value at a significance level of 0.05 and degrees of freedom (df) = n - 1 = 14 - 1 = 13.
Using a t-table or a statistical calculator, the critical t-value is approximately ±2.160.

Step 7: Make the decision.
If the absolute value of the calculated t-statistic is greater than the critical t-value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
|calculated t-statistic| = |(27.29 - 30) / (3.76 / √14)| = 1.976

Since 1.976 < 2.160, we fail to reject the null hypothesis.

Step 8: State the conclusion.
Based on the sample data and a 5% level of significance, there is not enough evidence to conclude that the mean delivery time is different from 30 minutes.

To determine if the mean delivery time is different than 30 minutes based on the given sample data, we can perform a hypothesis test.

The null hypothesis (H0) states that the mean delivery time is equal to 30 minutes, while the alternative hypothesis (H1) states that the mean delivery time is different from 30 minutes.

Let's go through the steps for hypothesis testing:

Step 1: Define the hypotheses:
- Null hypothesis (H0): The mean delivery time is equal to 30 minutes.
- Alternative hypothesis (H1): The mean delivery time is different from 30 minutes.

Step 2: Set the significance level (alpha):
The significance level (alpha) is given as 5% or 0.05.

Step 3: Calculate the test statistic:
Since we have the sample data, we can calculate the test statistic using the formula:
t = (sample mean - population mean) / (sample standard deviation / √n)

Step 4: Determine the critical value(s):
The critical value(s) can be obtained from the t-distribution table based on the degrees of freedom and the significance level (alpha).

Step 5: Compare the test statistic with the critical value(s):
If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Step 6: Draw the conclusion:
Based on the comparison in step 5, we can draw the conclusion whether to reject or fail to reject the null hypothesis.

Let's perform the calculations:

Step 3: Calculate the test statistic:
For the given sample data:
mean = (21 + 28 + 22 + 23 + 30 + 27 + 29 + 30 + 32 + 24 + 32 + 25 + 26 + 31) / 14 = 27.5
sample standard deviation = sqrt(((21-27.5)^2 + (28-27.5)^2 + ... + (31-27.5)^2) / (14-1)) = 3.57
n = 14
Thus, the test statistic is:
t = (27.5 - 30) / (3.57 / √14) ≈ -2.47

Step 4: Determine the critical value(s):
Since we have a two-tailed test and a significance level of 0.05, the critical value can be obtained from the t-distribution table with 13 degrees of freedom (n - 1 = 14 - 1 = 13):
The critical values are approximately -2.16 and 2.16.

Step 5: Compare the test statistic with the critical value(s):
The absolute value of the test statistic, 2.47, is larger than the critical value, 2.16.

Step 6: Draw the conclusion:
Since the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis. This means that based on the given sample data, there is evidence to suggest that the mean delivery time is different from 30 minutes.

In conclusion, the mean delivery time is significantly different from 30 minutes, based on the given data and using a 5% level of significance.

(27£¬14-30£©/(3.71/square root of 14) than find the p-value