Question

Directions: Please show your work. Submitting answers only will not earn you full credit for a given problem. Showing work includes, but is not limited to: formulas, graphs, and calculations. If you use the web calculator, StatCrunch, Minitab, Excel, Table V, Graphing Calculator, etc…, you must include your output with your answer (I suggest using the Snipping Tool in Windows). Use the prescribed method of test (Classical or P-value) for each question. Follow all the steps used in hypothesis testing as delineated in the guided notes or the textbook and be sure to verify the conditions for each test. Assume that all samples are randomly obtained.

Question 1: A fast-food restaurant manager wants to see the average bill per car passing through the drive-thru increase. Historically, cars spend an average of $12.38 with a standard deviation of $3.21.

a) After implementing the new process employees must follow for each car in the drive-thru, the manager randomly selects 40 cars and finds the mean bill is $12.94. Does this sample evidence suggest the new process results in a higher ticket price per car? Use the Classical approach with a 5% level of significance.

b) Based on your results of the hypothesis test—the decision you made in part a, what type of error might have occurred? Please explain.

Question 2: The quality control manager for a filling operation in a bottling plant is concerned with the variability in the volume of milk dispensed into gallon jugs. The filling process results in jugs whose volumes are normally distributed with a mean of 1.02 gallons. In order for this process to be in-control, the standard deviation should be less than 0.05 gallons (otherwise milk spills all over the production line). The manager takes a sample of 25 jugs of milk off the production line and determines the standard deviation to be 0.03 gallons. Is the process in-control with respect to the variability in the volume of milk dispensed into the jugs? Use the P-value approach with a 2.5% level of significance.

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u < 12.38
Alternative hypothesis: u > 12.38

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.50755
DF = n - 1

D.F = 39
t = (x - u) / SE

t = 1.103

t_critical = + 1.685

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 1.103.

Thus the P-value in this analysis is 0.138.

Interpret results. Since the P-value (0.138) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Interpret results. Since the t-value (1.103) does not lies in the rejection region, hence we cannot reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that the new process results in a higher ticket price per car.

b) Since we fail to reject the null hypothesis, hence there are chances that we might have committed type II error.