statistics and probability

Module 1 Homework Assignment.

Please analyze the following problems in JMP and answer the questions below.

1. In May 2008, CNN reported that Sports Utility Vehicles (SUVs) are plunging toward the “endangered” list. Due to soaring oil prices and environmental concerns, consumers are replacing gas-guzzling vehicles with fuel-efficient smaller cars. As a result, there has been a big drop in the demand for new and used SUVs. A sales manager of a used car dealership for SUVs believes that it takes more than 90 days, on average, to sell an SUV. In order to test his claim, he samples 40 recently sold SUVs and obtains the following data:

SUV 1	111
SUV 2	91
SUV 3	101
SUV 4	93
SUV 5	82
SUV 6	60
SUV 7	107
SUV 8	84
SUV 9	99
SUV 10	82
SUV 11	79
SUV 12	130
SUV 13	95
SUV 14	81
SUV 15	101
SUV 16	108
SUV 17	73
SUV 18	107
SUV 19	119
SUV 20	115
SUV 21	107
SUV 22	94
SUV 23	125
SUV 24	102
SUV 25	98
SUV 26	83
SUV 27	107
SUV 28	84
SUV 29	110
SUV 30	80
SUV 31	124
SUV 32	65
SUV 33	87
SUV 34	112
SUV 35	96
SUV 36	86
SUV 37	111
SUV 38	73
SUV 39	105
SUV 40	121

a) State the null and alternative hypothesis to test the sales manager’s belief.

here, we are to test,

Null hypothesis, 

Against Alternative hypothesis, 

b) What is the p-value for the hypothesis test of the sales manager’s belief?

Here 

= 97.2 and s= 16.87

So, the given test statistic is,

= 2.6993 

 2.70

Hence, the p-value of the test statistic is 0.0051

c) What is your conclusion about the sale’s manager’s claim? Write your answer in a sentence or two. Use α=0.05.

As, p-value is less than \alpha. We can say the result is significant. Hence we accept the manager's claim.

d) Interpret the 95% confidence interval for the number of days it takes to sell an SUV at this dealership.

The confidence interval is given by, 

i.e. 

= (91.8092, 102.5908) = (91.81, 102.59)

2. A bank with a branch located in a commercial district of a city has developed an improved process for serving customers during the noon- to- 1 P. M. lunch period. The waiting time (operationally defined as the time elapsed from when the customer enters the line until he or she reaches the teller window) needs to be shortened to increase customer satisfaction. A random sample of 15 customers is selected, and the results (in minutes) are as follows:

4.50 6.10 0.38 5.12 6.46 6.19 3.79 4.21

5.55 3.02 5.13 4.77 2.34 3.54 3.20

Suppose that another branch, located in a residential area, is also concerned with the noon- to-1pm lunch period. A random sample of 15 customers is selected, and the results are as follows:

10.49 6.68 5.64 4.08 6.17 9.91 5.47 9.66

5.90 8.02 5.79 8.73 3.82 8.01 8.35

Is there evidence of a difference in the mean waiting time between the two branches?

a. Describe the characteristics of the data collected from both populations (such as: Histogram, mean, median, variance, and standard deviation).

Variable	Mean	StDev	Variance	Median
Branch 1	4.287	1.638	2.683	4.500
Branch 2	7.115	2.082	4.336	6.680

b. Write out the Null and Alternative hypotheses to determine if there is evidence of a difference in the mean waiting time between the two branches.

c. Use the appropriate JMP analysis to draw a conclusion about the mean waiting time between the two branches. Write your answer in a few sentences and report the p-value of the test. Use α=0.05.

Two-Sample T-Test and CI: Branch 1, Branch 2

Method

μ₁: mean of Branch 1

µ₂: mean of Branch 2

Difference: μ₁ - µ₂

Equal variances are assumed for this analysis.

Descriptive Statistics

Sample	N	Mean	StDev	SE Mean
Branch 1	15	4.29	1.64	0.42
Branch 2	15	7.11	2.08	0.54

Estimation for Difference

Difference	Pooled StDev	95% CI for Difference
-2.828	1.873	(-4.229, -1.427)

Test

Null hypothesis	H₀: μ₁ - µ₂ = 0
Alternative hypothesis	H₁: μ₁ - µ₂ ≠ 0

T-Value	DF	P-Value
-4.13	28	0.000

Since p-value is less than alpha 0.05 we reject null hypothesis and conclude that there is a significant difference between the mean waiting times between branch 1 and branch2.

d. Interpret the 95% confidence interval for the mean waiting time difference between the two branches.

95% CI for Difference

(-4.229, -1.427)

Since zero is not included in the 95% CI we reject null hypothesis and conclude that there is a significant difference between the mean waiting times between branch 1 and branch2.