Determining the Statistical Significance

1426584307

 

 

Abstract

This paper will discuss the statistical analysis of tobacco use as well as self-rated health for 70 students evenly divided between male and female. The hypothesis test used for the tobacco analysis is a chi square test using two rows (male and female) and three columns (never use, sometimes use, and use often). The hypothesis test used for the self-rated health is a t-test comparison of means. Both of these tests are conducted using a percentile value of 0.05.

The chi square test associated with tobacco use resulted in a failure to reject the null hypothesis. This means that the probability of having the observed results due to random chance was greater than 0.05. The t-test for comparison of means associated with the self-rated health resulted in a rejection of the null hypothesis. This means that the probability of having the observed results due to random chance was less than 0.05. In both tests the calculations were carried out in an Excel file and will be presented in summary here.

The interpretation of the hypothesis test for tobacco use is that the observed data does not provide enough evidence that tobacco use is different between males and females at the 95% level of significance. On the other hand, the interpretation of the hypothesis test for self-rated health does suggest that males and females perceive their own health differently at the 95% level of significance. Relevant calculations will be provided in the appendix at the end of the report.

 

 

Determining the Statistical Significance for Tobacco Use

by Sex and Self-rated Health by Sex

 

This report presents the methodology and results for two different hypothesis tests performed on the same group of 70 students, 35 of whom were male and 35 of whom were female. The first test was concerned with determining whether or not tobacco use is different between male and female students. This was done using a chi square test that allowed for three possible responses for the frequency of the students’ tobacco use. The second test was concerned with determining whether or not there was a difference in how male and female students evaluated their own personal health. This was done using a t-test for comparison of means where the self-rating was on a scale from 0 to 10. The data for these tests came from surveys that were completed by the students. The results from the surveys were entered into an Excel file and then multiple filters were used to determine the information necessary for these determinations.

The first step for evaluating tobacco use was to set up the null hypothesis and the alternative hypothesis. As is typical with chi square contingency tables, the hypotheses associated with this particular test are

H0:      Tobacco use is independent of sex
HA:     Tobacco use is associated with sex.

Remember that this is a two tail test because we are only concerned with whether or not the two sexes use tobacco in different amounts, not which sex uses more or less tobacco.

The next step was to set up the contingency table. This was accomplished with the help of Excel and some filters that made counting simple and accurate. The completed contingency table is shown at the top of the following page.

Never Sometimes Often Total
Male 12 11 12 35
Female 17 14 4 35
Total 29 25 16 70

Notice that there were 35 male and 35 female respondents. Also notice that the bottom row and far right column both add to 70: the total number of student respondents.

Never Sometimes Often
Male 14.50 12.50 8.00
Female 14.50 12.50 8.00

Next the expected values for each possibility (six in total) were calculated. The resulting table is:

It is readily seen that the expected values are not equivalent to the observed values. The question that our hypothesis test will answer is whether or not that difference is significant at the five percent level.

The result of the test was a chi square value of 5.22. The degrees of freedom for this test was found by multiplying one less than the number of rows by one less than the number of columns. In other words this test has (2 – 1) * (3 – 1) = 2 degrees of freedom. The critical value for a chi square test with 2 degrees of freedom at the 0.05 level of significance is 5.99. Since our test statistic is lower than the critical value, corresponding to a p-value of 0.0745, we are unable to reject the null hypothesis.

A fairly similar order of steps, albeit a different technique, was used to analyze the individual health self-rating. As with the hypothesis test for tobacco use, the first step is to determine the null hypothesis and the alternative hypothesis. For the t-test comparison of means these hypotheses are:

H0:      The means are the same for males and females
HA:     The means are not the same for males and females

This is also a two tail test because we only want to know if there is a difference in the means, not which sex has a higher or lower self-rating.

The next step was to find both the mean and the standard deviation of the self-rating for males and for females. This is where the Excel file was most helpful, since it was able to count the value for each possible rating and calculate the needed values easily. For males we found that the average rating was 6.429 with a standard deviation of 1.899. For females we found that the average rating was 5.400 with a standard deviation of 1.866. Again the question of importance is whether or not this difference is significant at the five percent level.

In order to conduct the t-test for comparison of means, we needed to have the difference between the means, an estimate of variance for all 70 respondents, and an estimated standard error. Determining these led to a test statistic of 2.286. Since our hypothesis test had 68 degrees of freedom our critical value was 1.996. Here our test statistic is greater than the critical value, corresponding to a p-value of 0.0245; we conclude that the null hypothesis should be rejected.

This test lends itself to a graphical presentation fairly readily. Excel was again used, this time to produce frequency plots in the form of column tables for both males and females. The column table for males is

while the column table for females appears here:

Visually it appears that there are significantly lower ratings on average for females, which was confirmed at the 95 percent level by our hypothesis test. The distribution for males appears to be fairly normal, while the distribution for females seems to deviate from a normal distribution somewhat.

Both of these tests were conducted using data obtained from the same 70 student respondents. It is possible for these tests to be inconclusive, at least temporarily. In other words, if the critical value is 2.50 and the test statistic is 2.49, it can be difficult to say with certainty that the null hypothesis is unable to be rejected. When this happens, the best way to increase confidence in the results of the hypothesis test is to increase the sample size. For the hypothesis test regarding tobacco use this is not an issue. For the hypothesis test regarding self-rating of health the difference is smaller but not too small. One possible place where further testing might be useful is the female results for self-rating since the data here does not appear to be as normally distributed as that for the males.

 

 

Appendix for Tobacco

Never Sometimes Often Total
Male 12 11 12 35
Female 17 14 4 35
Total 29 25 16 70

Starting with the contingency table

the expected values were found using

(35 * 29) / 70 = 14.50                         Never use tobacco
(35 * 25) / 70 = 12.50                         Sometimes use tobacco
(35 * 16) / 70 = 8.00                           Often use tobacco.

The chi square statistic was found from
(12-14.5)2/14.5 + (17-14.5)2/14.5 + (11-12.5)2/12.5 + (14-12.5)2/12.5 +
(12 -8)2/8 + (4-8)2/8 = 5.2221

There are (3-1)*(2-1) = 2 degrees of freedom.

 

 

Appendix for self-rating

First find the mean rating for males

mean = [(2)(2) + (1)(3) + (5)(5) + (12)(6) + (7)(7) + (2)(8) + (4)(9) + (2)(10)] / 35
= 6.429,

and the men rating for females

mean = [(1)(2) + (6)(3) + (5)(4) + (5)(5) + (9)(6) + (5)(7) + (1)(8) + (3)(9)] / 35
= 5.400.

Next the Excel routine STDEV.S was used to find the standard deviations

SD for males = 1.899
SD for females = 1.866.

The difference in means is 6.429 – 5.400 = 1.029. The estimate of variance is

(1/2) * (1.8992 + 1.8662) = 3.544.

The estimated standard error is then

square root (2 * 3.544 / 35) = 0.450

Our test statistic is then t = 1.029 / 0.450 = 2.286.

Since there are (35 – 1) + (35 – 1) = 68 degrees of freedom, the critical value is 1.99.

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