MacDonald’s executives want to experiment with redesigning its restaurants so that the customers form one line leading to four registers, rather than four lines leading to four separate registers. They redesign 30 randomly selected restaurants with the single line. In addition, they randomly select 30 restaurants with the four line configuration to participate in the study. At each restaurant, an employee monitors the wait time (in minutes) of randomly selected patrons. Use the data below to test if the standard deviation for the wait time in the single line is less than the standard deviation of wait time in the multiple lines at the 0.0 5 level of significance.

Single Line Multiple Lines
1.2 2.1 1.7 2.8 1.1 1.6 2.9 4.6
1.9 2.1 2.4 3.0 3.8 2.9 2.8 2.9
2.1 2.3 1.9 2.9 4.3 2.3 2.0 2.6
2.7 2.0 2.9 2.3 1.3 1.3 2.0 2.7
2.8 1.1 3.1 1.8 2.0 3.2 2.3 0.9

To test if the standard deviation for the wait time in the single line is less than the standard deviation of wait time in the multiple lines at the 0.05 level of significance, we can follow these steps:

Step 1: State the hypotheses:
- Null Hypothesis (H0): The standard deviation of wait time in the single line is not less than the standard deviation of wait time in the multiple lines.
- Alternative Hypothesis (Ha): The standard deviation of wait time in the single line is less than the standard deviation of wait time in the multiple lines.

Step 2: Determine the level of significance (α):
The level of significance is given as 0.05 in the problem.

Step 3: Compute the test statistic:
We will use the F-test to compare the variances of the two samples. However, since we are only interested in testing if the standard deviation of the wait time in the single line is less than the standard deviation of the wait time in the multiple lines, we can compare the sample variances directly without calculating the F-test.

To calculate the sample standard deviation for each group (single line and multiple lines), follow these steps:

For the Single Line group:
- Calculate the mean (x̄) for the wait time.
- Subtract the mean from each wait time value, square the result, and sum up all the squared differences.
- Divide the sum by (n-1), where n is the number of observations in the single line group.
- Take the square root of the result to get the sample standard deviation.

For the Multiple Lines group:
- Follow the same steps as above using the wait time values for the multiple lines.

The calculation for the standard deviations is as follows:

Single Line:
- x̄ = (1.2 + 2.1 + 1.7 + 2.8 + 1.9 + 2.1 + 2.4 + 3.0 + 2.1 + 2.3 + 1.9 + 2.9 + 2.7 + 2.0 + 2.9 + 2.3 + 2.8 + 1.1 + 3.1 + 1.8) / 20 = 2.275
- Sum of squared differences = (1.2 - 2.275)² + (2.1 - 2.275)² + ... + (1.8 - 2.275)² = 9.775
- Sample standard deviation = √(9.775 / 19) = 0.9298

Multiple Lines:
- x̄ = (1.1 + 1.6 + 2.9 + 4.6 + 3.8 + 2.9 + 2.8 + 2.9 + 4.3 + 2.3 + 2.0 + 2.6 + 1.3 + 1.3 + 2.0 + 2.7 + 2.0 + 3.2 + 2.3 + 0.9) / 20 = 2.415
- Sum of squared differences = (1.1 - 2.415)² + (1.6 - 2.415)² + ... + (0.9 - 2.415)² = 16.6925
- Sample standard deviation = √(16.6925 / 19) = 1.3419

Step 4: Compare the sample standard deviations:
- We will compare the sample standard deviations directly since we have determined that the standard deviation of the single line should be less than the standard deviation of multiple lines in the alternative hypothesis (Ha).
- 0.9298 < 1.3419

Step 5: Make a decision:
Since the sample standard deviation of the single line (0.9298) is less than the sample standard deviation of the multiple lines (1.3419), we reject the null hypothesis (H0) and conclude that there is evidence to support the claim that the standard deviation for the wait time in the single line is less than the standard deviation of wait time in the multiple lines.

Note: If you wanted to calculate the test statistic and compare using the F-test, you would calculate the F statistic as the ratio of the sample variances and compare it against the critical F value at the given level of significance. However, in this case, comparing the sample standard deviations directly provides the same conclusion.

To test if the standard deviation for the wait time in the single line is less than the standard deviation of the wait time in the multiple lines, we can perform a hypothesis test.

Let's define our null and alternative hypotheses:

Null hypothesis (H0): The standard deviation for the wait time in the single line is not less than the standard deviation of wait time in the multiple lines.
Alternative hypothesis (Ha): The standard deviation for the wait time in the single line is less than the standard deviation of wait time in the multiple lines.

To test this hypothesis, we can use the F-test, which compares the variances of two populations.

Here are the steps to perform the hypothesis test using the given data:

Step 1: Calculate the sample standard deviations for each set of data (single line and multiple lines).

For the single line data:
Standard deviation of single line (s1) = 0.728

For the multiple lines data:
Standard deviation of multiple lines (s2) = 0.963

Step 2: Set up the F-test. Calculate the F-statistic.

F-statistic = (s1^2) / (s2^2)

F-statistic = (0.728^2) / (0.963^2) = 0.589

Step 3: Determine the degrees of freedom for the F-test.

Degrees of freedom for the numerator (single line) = n1 - 1 = 30 - 1 = 29
Degrees of freedom for the denominator (multiple lines) = n2 - 1 = 30 - 1 = 29

Step 4: Look up the critical F-value for a significance level of 0.05 and the calculated degrees of freedom.

Using a table or statistical software, we find the critical F-value to be approximately 0.706.

Step 5: Compare the calculated F-statistic to the critical F-value.

Since the calculated F-statistic (0.589) is less than the critical F-value (0.706), we fail to reject the null hypothesis.

Step 6: Interpret the results.

Based on our analysis, there is not enough evidence to conclude that the standard deviation for the wait time in the single line is less than the standard deviation of wait time in the multiple lines at the 0.05 level of significance.