Quality Associates Inc.

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Quality Associates Inc., a consultancy company aims to offer sampling and statistical information to a customer. The mean 12 and the standard 0.21 variance for a sample of the satisfactory client method are calculated. Successive samples of 30 findings are collected at random to assess the customers’ satisfaction level. The method of the client was deemed successful if the average was 12. If the mean was substantially different for this process from 12, then the process would be unsatisfactory and corrective steps needed to restore it to the level of organizational satisfaction. The following table shows the descriptive statistics for the four successive samples obtained for the process.

SUMMARY

Groups

Count

Sum

Average

Variance

Standard deviation

Sample 1

30

358.76

11.95867

0.048557

0.220356

Sample 2

30

360.86

12.02867

0.048557

0.220356

Sample 3

30

356.67

11.889

0.04292

0.207171

Sample 4

30

362.44

12.08133

0.042481

0.206109

The process mean for sample 1 is 11.96, sample 2 is 12.03, sample 3 is 11.89 and sample 4 is 12.08. The standard deviation for sample 1 and sample 2 is 0.22 while the standard deviation for sample 3 and sample 4 is 0.21. The study assumes that the four standard deviations are equal. The assumption of the equality of variances appears reasonable because the standard deviations for sample 3 and 4 are equal to the population standard deviation while the standard deviation for sample 1 and 2 have a small difference from the population standard deviation.

Method

The study will use inferential statistics to evaluate the satisfaction of the client’s process. Sample t-tests will be used to determine the significance of the difference between the population mean and the sample mean. Microsoft Excel software will be used to analyze the data.

Results

The significance of the difference between the each sample mean and the population mean will be determined. The table below shows the results of the independent sample t-test for the difference between mean for sample 1 and population mean.

One-Sample Test

Test Value = 12

t

df

Sig. (2-tailed)

Mean Difference

99% Confidence Interval of the Difference

Lower

Upper

Sample 1

-1.027

29

.313

-.04133

-.1522

.0696

The process mean for sample 1 is significantly equal to the population mean, 12. This is because the p-value, 0.313 is greater than the level of significance, 1%. Therefore, the null hypothesis is not rejected.

The table below shows the t-test results for sample 2.

One-Sample Test

Test Value = 12

t

df

Sig. (2-tailed)

Mean Difference

99% Confidence Interval of the Difference

Lower

Upper

Sample 2

.713

29

.482

.02867

-.0822

.1396

The p-value, 0.482 is greater than the level of significance, 1%. The test statistic is not significant. Therefore, the null hypothesis is not rejected. The process mean for sample 2 is significantly equal to the population mean.

The table below shows the result for sample t-test on sample 3

One-Sample Test

Test Value = 12

t

df

Sig. (2-tailed)

Mean Difference

99% Confidence Interval of the Difference

Lower

Upper

Sample 3

-2.935

29

.006

-.11100

-.2153

-.0067

The test statistic is significant because the p-value, 0.006 is lower than the level of significance, 1%. Therefore, the null hypothesis is rejected. The process mean for sample 3 is different from the population mean.

The table below shows the t-test results for the mean of process 4.

One-Sample Test

Test Value = 12

t

df

Sig. (2-tailed)

Mean Difference

99% Confidence Interval of the Difference

Lower

Upper

Sample 4

2.161

29

.039

.08133

-.0224

.1851

The test statistic is insignificant because the p-value, 0.039 is greater than the level of significance, 1%. Therefore, the null hypothesis is not rejected. The mean process mean for sample 4 is significantly equal to the population process mean.

Further, we wish to determine whether there exists significant differences among the sample process means for the four samples. The table below shows the results of the analysis of variance (ANOVA) to determine the significance of the difference among the groups.

ANOVA

Value

Sum of Squares

df

Mean Square

F

Sig.

Between Groups

.631

3

.210

4.606

.004

Within Groups

5.293

116

.046

Total

5.923

119

The test statistic, F = 4.606 is significant because the p-value, 0.004 is less than the level of significance, 1%. Therefore, the null hypothesis is rejected. We, therefore, conclude that at least one of the sample means is significantly different from the others.

The table below shows the comparison between the sample means.

Value

Tukey B

Sample

N

Subset for alpha = 0.05

1

2

Sample 3

30

11.8890

Sample 1

30

11.9587

11.9587

Sample 2

30

12.0287

Sample 4

30

12.0813

Means for groups in homogeneous subsets are displayed.

a. Uses Harmonic Mean Sample Size = 30.000.

The sample mean for sample 3 and sample 1 are equal. Sample means for sample 1, sample 2 and sample 4 are also significantly equal. However, the sample mean for sample 3 is significantly different from the sample mean for sample 2 and sample 4.

The table below shows the confidence intervals for the sample means.

Sample 1

 

Sample 2

 

Sample 3

 

Sample 4

 

Confidence Level(99.0%)

0.11089

Confidence Level(99.0%)

0.11089

Confidence Level(99.0%)

0.10426

Confidence Level(99.0%)

0.10372

Lower limit

11.8478

Lower limit

11.9178

Lower limit

11.7847

Lower limit

11.9776

Upper limit

12.0696

Upper limit

12.1396

Upper limit

11.9933

Upper limit

12.1851

The confidence intervals for sample 1, sample 2 and sample 4 are inclusive of 12. This implies that the sample means are significantly equal to 12. However, the confidence interval for sample 3 is different from 12.

Conclusion

The sample t-test shows that all the sample means are equal to 12 except the mean for sample 3. Therefore, there is a need for corrective action after the time when sample 3 was obtained. The corrective actions would be necessary to increase mean of the process to 12 since the sample mean was less than 12.

The results of the test could be affected by choice of the significance level. When the significance level is increased, the acceptance region for the study is reduced (Robert, 2013). This in effect reduces the probability of rejecting the null hypothesis. Thus, the increase in the significance level increases the probability of type I error and reduces the probability of type II error (Murray & Larry, 2014)

References

Murray, S. R., & Larry, S. J. (2014). Statistics. New York: McGraw-Hill.

Robert, D. J. (2013). Business statistics. Boston: Pearson.

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