Question

- Problem Par, Inc. Par, Inc., is a major manufacturer of golf equipment Management believe share could be increased with the introduction of a cut-resistant, longer-lasting golf ball. There- fore, the research group at Par has been investigating a new golf ball coating designed to resist cuts and provide a more durable ball. The tests with the coating have been promising.

excel will be helpful too

Answer 1

Answer:

1) Here we have to test the two model means( current and new):

under this test the null and alternative hypothesis are,

$_{Ho}$: There is no significant difference between the means.

$_{Ha}$: There is significance difference between the means.

let us assume the level of significance $\alpha$ = 0.05

2)

use minitab to get the required results:

1. import the data into the minitab worksheet.

2. go to start $\rightarrow$ basic statics$\rightarrow$2 sample t.

3. Specify the samples in the different columns and click on options.

4.Specify the required values and click OK.

5. Click ok.

By following the above instructions we get the following output.

Two sample T-Test and CI: Current , New\

Two - sample T for Current vs New

N Mean StDev SE Mean

Current 40 270.27 8.75 1.4

New 40 267.50 9.90 1.6

Difference = mu (Current) - mu (New)

Estimate for difference: 2.77

95% CI for difference: (-1.39, 6.94)

T-Test of difference = 0 (vs not =):

T-Value = 1.33 P-Value = 0.188 DF = 76

From the above output we have the test statistic value as t = 1.33 and the corresponding P. value is 0.188, which is greater than the given level of significance, so we fail to reject the null hypothesis and conclude that there is a significance difference between the two mean values.

3)

Use Excel to get the required results:

Descriptive statistics

count 40 40

mean 270.28 267.50

sample variance 76.61 97.95

sample standard deviation 8.75 9.90

minimum 255 250

maximum 289 289

median 270.00 265.00

mode 272.00 263.00

From the above output we observe that the mean for Current data is 270.28 and for the new data is 267.50. The variance value for the current data is 76.61 and for the new data is 97.95

4).

here we need to find the 95%confidence interval for the both the data sets.

Use MINITAB we get the required results.

1. Import the data into the MINITAB work sheet .

2. Go to Stat$\rightarrow$Basic statistics $\rightarrow$ Isample.

3.Specify the samples in the different columns and click on options

4. specify the required values and click ok.

5. click ok.

BY following the above instructions we get the following output

One-Sample T: Current, New Variable Current New 40 40 Mean St Dev 270.27 8.75 267.50 9.90 SE Mean 1.38 1.56 95% CI (267.48, 273.07) (264.33, 270.67)

From the above output we have the 95%confidence interval for current data is

(267.48 , 273.07 ) and for the new data is (264.33 , 270.67)

now ,we need to find the 95% confidence interval for the difference of means

Use MINITAB to get the required results:

1. Import the data into the MINITAB work sheet.

2. Go to Stat$\rightarrow$Basic statistcs$\rightarrow$2 sample t.

3. specify the samples in the different columns and click on options.

4. Specify the required values and click OK.

5.Click ok.

By following the above instructions we get the following output.

Two sample T-Test and CI : Current, New

Two - sample T for Current vs New

N Mean stDev SE Mean

Current 40 270.27 8.75 1.4

New 40 267.50 9.90 1.6

Difference = mu (current ) - mu (New)

Estimate for difference :2.77

95%CI for difference : (-1.39, 6.94)

T-Test of difference = 0 (vs not =):

T-Teset = 1.333 p-value = 0.188 DF = 76

From the above output we have the 95% confidence interval for differnce of means is (-1.39,6.94)

5). There is no need to increase the sample for the given problem since the given data provides the sufficient evidence to compare the driving distance from two types of bails.