The Jefferson valley bank claims that the variation of their wait time is less when using a single line that feeds three tellers as opposed to having threes peerage lines. Test the claim that the standard deviation for wait time is less than 1.9 minutes with a single feed line at alpha=0.05 for the following sample of customers wait times in minutes and write your conclusion. Values are 6.5 6.6 6.7 6.8 7.1 7.3 7.4 7.7. 7.7 7.7

To test the claim made by the Jefferson valley bank, we need to perform a hypothesis test using the provided sample data. Here's how you can conduct the hypothesis test to determine if the standard deviation for wait time is less than 1.9 minutes:

Step 1: State the hypotheses:
- Null Hypothesis (H0): The standard deviation for wait time is not less than 1.9 minutes.
- Alternative Hypothesis (H1): The standard deviation for wait time is less than 1.9 minutes.

Step 2: Set the significance level (alpha) at 0.05, which is given in the question.

Step 3: Calculate the sample standard deviation. In this case, the sample data is provided as follows: 6.5, 6.6, 6.7, 6.8, 7.1, 7.3, 7.4, 7.7, 7.7, 7.7. To find the sample standard deviation, follow these steps:
a. Find the mean (average) of the data set: (6.5 + 6.6 + 6.7 + 6.8 + 7.1 + 7.3 + 7.4 + 7.7 + 7.7 + 7.7) / 10 = 7.05.
b. Calculate the deviation from the mean for each data point: subtract the mean from each data point.
e.g., Deviation of 6.5 from the mean: 6.5 - 7.05 = -0.55.
c. Square each deviation: (-0.55)^2, (-0.45)^2, (-0.35)^2, ..., (0.65)^2, (0.65)^2, (0.65)^2.
d. Find the sum of squared deviations: sum up all the squared deviations obtained in the previous step.
e. Divide the sum of squared deviations by (n-1), where n is the number of data points. In this case, n is 10.
f. Take the square root of the result obtained in step (e). This will give us the sample standard deviation.

Step 4: Conduct the hypothesis test using a chi-square test statistic.
a. The test statistic for this test is [ (n-1) * sample standard deviation^2 ] / population standard deviation^2.
Since we don't have the population standard deviation, we will use the hypothesized value of 1.9 minutes as the population standard deviation.
b. Plug in the values into the formula to calculate the test statistic.

Step 5: Determine the critical value or p-value.
a. Since the alternative hypothesis is one-tailed (less than), we will use the lower tail of the chi-square distribution.
b. Look up the critical value for the given alpha (0.05) and the degrees of freedom (n-1, which is 10-1 = 9) in the chi-square distribution table, or use a statistical calculator.
Note: The degrees of freedom are equal to the number of data points minus 1.
c. Alternatively, you can calculate the p-value using the chi-square test statistic. The p-value represents the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true.

Step 6: Make the decision and state the conclusion.
a. If the test statistic is less than the critical value or the p-value is less than the alpha level, reject the null hypothesis.
b. If the test statistic is greater than the critical value or the p-value is greater than the alpha level, fail to reject the null hypothesis.

By following these steps, you can test the claim made by the Jefferson valley bank and arrive at a conclusion based on the hypothesis test.