Assignment -No Plagiarism PLEASE read before accepting assignment. Assignment is Due Sunday, October 18, 2015 by 1pm EST

10000200003000040000

95% Confidence Interval for Mu

250002600027000280002900030000310003200033000

95% Confidence Interval for Median

Variable: June Sales

A-Squared:

P-Value:

Mean

StDev

Variance

Skewness

Kurtosis

N

Minimum

1st Quartile

Median

3rd Quartile

Maximum

25897.5

5582.3

25434.0

0.348

0.444

29218.8

7296.5

53239230

-4.5E-01

0.154906

21

12345.0

23888.5

31232.0

34878.5

43243.0

32540.1

10536.7

33065.6

Anderson-Darling Normality Test

95% Confidence Interval for Mu

95% Confidence Interval for Sigma

95% Confidence Interval for Median

Descriptive Statistics

Excel Graph

May and June Sales

0

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10000

15000

20000

25000

30000

35000

40000

45000

50000

123456789101112131415161718192021

Sales Each week

Dollar Value

Sales in MaySales in June

Hypothesis Test

Hypothesis

HO; the mean of the sales for June are less than or equal to May sales

H1: The mean for June are greater than the mean sale for May.

Decision Rule: If the probability value calculated based upon the sample is less than the significance level of .05, I will reject the null hypothesis and conclude the sales of the two months are equal

P Value = 0.036

Decision: Since the probability value of .036 is less than .05, I will reject the null hypothesis and conclude the sales in June are higher than May. However, since the sales figures for only one month, they would not constitute a trend because of the limited time frame of the data. However, they would seem to indicate that longer test marketing would be justified because the probability value of the test sample of .036 is very low and therefore the probability that these results are due to just random variation is very small. If this figure was closer to the .05 level, I would be less confident that the results are not due to random variation. Additionally, a longer time for collection of data, with a year being recommend, would allow for analysis of any seasonal trends and also the greater the sample size, the data would reduce the potential variation because it would provide for greater accuracy in the results.

Hypothesis Testing Paired T

You are in charge of new product development for your company’s electronic division. One of the key aspects of electronics is the ability for the battery to extend the time on the device, specifically battery power. You are considering two batteries, and old standard battery of a new one that supposedly has a longer life. You want to test each set of batteries in the same computer with the same people testing them to minimize any outside influences in the test. You want to be very stringent in your analysis so you choose the .01 level of significance. Is there any difference between the two?

HO:

H1

Decision Rule

Test Statistic:

Decision

Notes:

Participant	New Battery	Old Battery
Bob	45	52
May	41	34
Deno	53	40
Sri	40	38
Pat	43	38
Alexis	43	44
Scott	49	34
Aretha	39	45
Jen	41	28
Ben	43	33

Answer

You are in charge of new product development for your company’s electronic division. One of the key aspects of electronics is the ability for the battery to extend the time on the device, specifically battery power. You are considering two batteries, and old standard battery of a new one that supposedly has a longer life. You want to test each set of batteries in the same computer with the same people testing them to minimize any outside influences in the test. You want to be very stringent in your analysis so you choose the .01 level of significance. Is there any difference between the two?

HO: Battery life of old is equal to battery life of new

H1 Battery life of old is not equal to battery life of new

Decision Rule

Decision Rule: If the probability value calculated is less than .01, we will reject the HO

Test Statistic: .073

Decision: Since the probability value calculated of .073 is greater than the critical value of .01, we fail to reject HO and are somewhat confident the battery lives are equal.

Paired T-Test and CI: New Battery, Old Battery

Paired T for New Battery - Old Battery

N Mean StDev SE Mean

New Battery 10 43.70 4.32 1.37

Old Battery 10 38.60 6.98 2.21

Difference 10 5.10 7.94 2.51

95% CI for mean difference: (-0.58, 10.78)

T-Test of mean difference = 0 (vs not = 0): T-Value = 2.03 P-Value = 0.073

Notes:

Late Fee versus No Late Fee

A blockbuster rental is becoming worried that with the advent of Netflix etc that customer defection to these new online rentals may decrease profits. Specifically, customers have complained about late fees and thus the local managers want to know if there new policy of no late fees has increased movie rentals. At the .10 level of significance, what are your findings?

HO:

H1

Decision Rule

Test Statistic:

Decision:

Customer	No Late Fee	Late Fee
1	14	10
2	12	7
3	14	10
4	13	13
5	10	9
6	13	14
7	12	12
8	10	7
9	13	13
10	13	9

Late Fee versus No Late Fee

A blockbuster rental is becoming worried that with the advent of Netflix etc that customer defection to these new online rentals may decrease profits. Specifically, customers have complained about late fees and thus the local managers want to know if there new policy of no late fees has increased movie rentals. At the .10 level of significance, what are your findings?

HO: No Late Fee Movie rentals have decreased or are equal to Late Fee Rentals

H1 No Late Fee Movie rentals have increased over Late Fee Rentals

Decision Rule If the probability value calculated is less than .10, we will reject the HO

Test Statistic: .009

Decision: Since the p calculated of .000 is less than .10, we reject HO and conclude there is strong evidence to conclude movie rentals have increased.

Paired T-Test and CI: No Late Fee, Late Fee

Paired T for No Late Fee - Late Fee

N Mean StDev SE Mean

No Late Fee 10 12.400 1.430 0.452

Late Fee 10 10.400 2.503 0.792

Difference 10 2.000 2.211 0.699

90% lower bound for mean difference: 1.033

T-Test of mean difference = 0 (vs > 0): T-Value = 2.86 P-Value = 0.009

Notes:

Minitab Instructions

Comparing Two Means with Minitab

I am comparing the mileage of two gasoline’s to see if Chevron exceeds Shell gasoline in mileage test (DATA IS NOT REAL). I am interested in is there a difference in the gas mileage between the two brands.

Step 1

This is not a paired T test rather a standard two sample T test.

Step 2

Since the data is in two columns, click on the Samples in Different Columns options and then click to enter the two columns of data for analysis In this case, Chevron was picked and Shell

Step 3

You have a choice of two graphs and it is personal preference if you want to use either graph.

Step 4

You choose your alternative hypothesis and also the level of significance as indicated by the level of confidence. For example, a .05 level of significance corresponds to a 95% confidence level.

Step 5

Evaluate your graphics

Step 6:

Final Results as seen in Minitab

Two-sample T for Chevron vs Shell

N Mean StDev SE Mean

Chevron 15 29.73 4.92 1.3

Shell 15 25.00 3.95 1.0

Difference = mu (Chevron) - mu (Shell)

Estimate for difference: 4.73333

95% CI for difference: (1.39747, 8.06920)

T-Test of difference = 0 (vs not =): T-Value = 2.91 P-Value = 0.007 DF = 28

Both use Pooled StDev = 4.4599

Arrows indicate how to read print out for the H1 is Chevron is not equal to Shell

Pay close attention to the sequence listed with the green being first or like a green light start and red is the second half of the H1 and red being like a red stop light. (

Two Sample T Test Using Summarized Data

We can use a two sample t test to evaluate how well people do on learning this can be used to evaluate training etc by administering the same test to two different groups, one with training and one without training to see if the training will pay off. In this case, we will simply call them sample one (received training) and sample two (did not receive training) and then we gave them the same test

The H 1 choices are less than, not equal to or greater than.

In this example, we are assuming that the first sample mean is greater than the second sample mean so we choose greater than.

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean

1 20 38.00 3.00 0.67

2 30 35.00 5.00 0.91

Difference = mu (1) - mu (2)

Estimate for difference: 3.00

95% lower bound for difference: 0.91

T-Test of difference = 0 (vs >): T-Value = 2.41 P-Value = 0.010 DF = 48

Both use Pooled StDev = 4.3205

HO: Sample one is less than or equal to sample two

H1: sample one is greater than sample two

Decision rule: If P calculated is less than .05, reject HO

Test stat: P value of .010

Decision: Since the p value calculated of .010 less than the critical value of .05, we reject HO and conclude sample group one has a higher training score than group two. For if the HO was true that the training group one has the same or lower score than training group two, the probability that you would have randomly found a sample averages of 38 and 35 is is 1.0%, therefore, it is possible not probable.

Sample Mean

Level of significance/ confidence level

Alternative Hypothesis

Graph Choices

Two columns selected

Select 2 Sample T

Enter summarized data

Click on summarized data

Choose H1