N Sample Size The Variances example essay topic

The unit of observations were random samples of twenty-five various employees divided into two distinct, independent populations, smokers and non-smokers. Then data on their absences from work for the previous year were obtained and used in this statistical inference. Because of a strong association between smoking and ill-health, it is generally accepted that smokers miss more work than their non-smoking counterparts. Does the smoker miss more work than the non-smoker? Data from these random samples were used to draw a conclusion... Unit of Observation: Smoking and Non-Smoking Employees Variable Name: Definition: Unit of Measurement: Data Source: Smoker Employee Days Absent in Past Year web Smokes July 11, 2000 Non-Smoker Employee Does Days Absent in Past Year web NOT Smoke July 11, 2000 RANDOM DATA SAMPLES LISTING Smokers: Absentees: Non-Smokers: Absentees: Employee 1 10 Employee 1 5 Employee 2 8 Employee 2 9 Employee 3 18 Employee 3 2 Employee 4 8 Employee 4 10 Employee 5 11 Employee 5 12 Employee 6 17 Employee 6 11 Employee 7 19 Employee 7 6 Employee 8 21 Employee 8 9 Employee 9 16 Employee 9 12 Employee 10 2 Employee 10 8 Employee 11 4 Employee 11 4 Employee 12 12 Employee 12 7 Employee 13 11 Employee 13 13 Employee 14 6 Employee 14 6 Employee 15 9 Employee 15 7 Employee 16 13 Employee 16 11 Employee 17 24 Employee 17 10 Employee 18 15 Employee 18 18 Employee 19 14 Employee 19 20 Employee 20 3 Employee 20 4 Employee 21 0 Employee 21 10 Employee 22 9 Employee 22 2 Employee 23 11 Employee 23 8 Employee 24 19 Employee 24 5 Employee 25 10 Employee 25 10 Mean: 11.6 Mean: 8.76 Standard Deviation: 6.110100927 Standard Deviation: 4.352011029 Variances: 37.33333333 Variances: 18.94 SOURCE: web July 11, 2000 STATISTICAL ANALYSISOUTPUTF-Test Two-Sample for Variances Smokers Non-SmokersMean 11.6 8.76 Standard Deviation 6.110100927 4.352011029 Variance 37.33333333 18.94 Observations 25 25 df 24 24 F 1.971136924 P (F = f) one-tail 0.051571253 F Critical one-tail 1.983757159 t-Test: Two-Sample Assuming Equal Variances Smokers Non-SmokersMean 11.6 8.76 Variance 37.33333333 18.94 Observations 25 25 Pooled Variance 28.13666667 Hypothesized Mean Difference 0 df 48 t Stat 1.892940764 P (T = t) one-tail 0.032201762 t Critical one-tail 1.677224191 P (T = t) two-tail 0.064403523 t Critical two-tail 2.01063358 CAN WE ACCEPT THE STATISTICAL NULL HYPOTHESIS ANALYSIS The data source used in this inference was found through a search engine, web After hours of surfing the web and grueling through mounds of data I used two of the random samples found at web For this statistical inference, the question was whether the means were truly different or could they have been samples from the same population.

To do draw a conclusion, we must first assume normal distribution. We must also set the null hypothesis to m 1 - m 2 = 0. And per this assignment we must set the a-level at. 05 and the hypothesis alternative to m 1 - m 2 ^1 0; thus requiring a two-tailed test. The random samples have a mean of 11.6 days absent for the smoker and 8.76 days absent for the non-smoker. All of my calculations were done using the data analysis tool in Excel but can be done manually with given equations: Sample Mean : n = sample size The variances of each sample are 37.33333333 for the smoking population and 18.94 for the non-smoking population.

Their standard deviations are 6.110100927 and 4.352011029 respectively. Manually: Sample Variance (s 2): Std Dev (s): Next we must determine if the population variances of these two samples are equal at an a-level of. 05. Based on the data and the assumption that the populations are normal, the variances are equal though barely (Fc = 1.983757159 and the F-test = 1.971136824). (Note if this was tested at a different a-level, we may not of accepted the variances as equal and would use a different equation to test the hypothesis). With the use of the F-table, these variances can be tested for equality manually also: To determine if Variances are equal : n 1 = n 1 - 1 Use F table: n 2 = n 2 - 1 Reject Region (Fc): F Fa, n 1, n 2 Test Statistic: Since the variances were tested and found equal at the a-level of.

05. Then we must use the pooled variances (the weighted average of the two sample variances) to calculate the test statistic. Excel did this calculation for me: 28.136666667. It can be done manually using this equation: Pooled variance = Now we perform the t-test under the assumption that is normally distributed. Again, this can be done manually: tc = ta/2, n 1+n 2-2 Utilizing the data analysis tool in Excel, I found the t-statistic for these sample means is 1.892940764 and the tc for a two-tailed test is +/- 2.01063358. Since the t-statistic falls within this range we accept the null hypothesis that the means of the samples are equal.

Based on these samples, the data does not offer evidence of differences in absentees in the workplace between smokers and non-smokers. I would like to note though that the test's conclusion very much depended on the a-significance level. If we had set the hypothesis alternative to m 1 - m 2 0 and did a one-tailed test, the tc would be 1.677224191. Since the Reject Region is t tc, in this instance, we would reject the null hypothesis and thus conclude that the mean of absentees in smoking workers is greater than the mean of absentees in non-smoking workers. I feel the results of this test should not be used as a basis for any decisions or policies. Further studies with greater number of samples would be needed to assess the true means of the populations..