. use W:\WP51\Biometry\AAAABiostat1726Spring2009\Homework\HW3\hw3old.dta

Transcription

1 . use W:\WP51\Biometry\AAAABiostat1726Spring2009\Homework\HW3\hw3old.dta. des Contains data from W:\WP51\Biometry\AAAABiostat1726Spring2009\Homework\HW3\hw3old.dta obs: 147 vars: 8 28 Jan :28 size: 5,292 (99.9% of memory free) storage display value variable name type format label variable label id float %9.0g lungca float %9.0g patcontlbl Lung cancer - patient/control sex float %9.0g sexlbl Sex socio float %9.0g sociolbl Socioeconomic status bird float %9.0g yesno Keeps birds - yes/no age float %9.0g Age in years year float %9.0g Years of smoking prior to diagnosis or examination smokrate float %9.0g Average rate of smoking in cigarettes per day Sorted by: Step 1: The first thing I would do is to use codebook to get an idea of what kind of variables we have and whether we have any missing data. Below we can see there is no missing data, that lung cancer, bird-keeping and socioeconomic status are all categorical variables with exactly two categories and that years of smoking appears to be a continuous variable.. codebook lung bird socio year lungca Lung cancer - patient/control label: patcontlbl range: [0,1] units: 1 unique values: 2 missing.: 0/147 tabulation: Freq. Numeric Label 98 0 Control 49 1 Patient bird Keeps birds - yes/no label: yesno range: [0,1] units: 1 unique values: 2 missing.: 0/147 tabulation: Freq. Numeric Label 80 0 No 67 1 Yes

2 socio Socioeconomic status label: sociolbl range: [0,1] units: 1 unique values: 2 missing.: 0/147 tabulation: Freq. Numeric Label Low 45 1 High year range: [0,50] units: 1 unique values: 40 missing.: 0/147 mean: std. dev: percentiles: 10% 25% 50% 75% 90% Step 2: Since year seems to be a continuous variable I have looked at sum(year), det and graphs (a histogram and a box plot). We can see that the skewness = (the histogram is skewed to the left) and that the kurtosis is 2.48 (the histogram is flatter than the normal distribution). My choice would be to run both the t-test (because the t-test is very robust to non-normality) and the Wilcoxon ranksum test (the non-parametric version of the t-test). If the answers are similar I will report the t-test because more people will recognize the t-test than will recognize Wilcoxon s ranksum test.. sum(year),det Percentiles Smallest 1% 0 0 5% % 0 0 Obs % 20 0 Sum of Wgt % 30 Mean Largest Std. Dev % % Variance % Skewness % Kurtosis

3 Frequency Step 3: The results of the two-tailed t-test are t = -3.70, df = 145 and p = The results of the ranksum test are z = and p = From both analyses we conclude that the lung cancer patients and the controls are different with respect to the number of years of smoking (the lung cancer patients (33.6 years) have smoked longer than the controls (25.0 years)). I would report the t-test here. We looked at the t-test and ranksum tests because the lung cancer variable had 2 categories and the years of smoking was continuous.. ttest year,by(lung) Two-sample t test with equal variances Group Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] Control Patient combined diff diff = mean(control) - mean(patient) t = Ho: diff = 0 degrees of freedom = 145 Ha: diff < 0 Ha: diff!= 0 Ha: diff > 0 Pr(T < t) = Pr( T > t ) = Pr(T > t) = ranksum year,by(lung)

4 Two-sample Wilcoxon rank-sum (Mann-Whitney) test lungca obs rank sum expected Control Patient combined unadjusted variance adjustment for ties adjusted variance Ho: year(lungca==control) = year(lungca==patient) z = Prob > z = Step 4: To consider bird-keeping versus cases and controls. So we have two categorical variables each with two categories. We first check to see if the expected values are large enough and they are (all expected values are larger than 5). We can use either the tabulate command or the case/control command (cc). The two printouts look quite different but pretty much provide the same data except tab gives you the expected values and cc gives you the odds ratio. Both indicate that 34.7% of the controls are bird keepers and 67.4 percent of the cancer cases keep birds. These percentages are statistically different (chi square = 14.04, df = 1, p < 0.001). The odds ratio indicates that the cases are 4 times more likely to keep birds than the controls.. tab lung bird,row exp chi2 Key frequency expected frequency row percentage Lung cancer - patient/co Keeps birds - yes/no ntrol No Yes Total Control Patient Total Pearson chi2(1) = Pr = 0.000

5 . cc lung bird Proportion Exposed Unexposed Total Exposed Cases Controls Total Point estimate [95% Conf. Interval] Odds ratio (exact) Attr. frac. ex (exact) Attr. frac. pop chi2(1) = Pr>chi2 = Step 5: In order to look at the socioeconomic status of the cases and controls I have again used both the tab and the cc commands. Using the tab command we get the expected values and see that they are all larger than 5 so it is ok to use the chi square test for this data. Using the chi square test we get that the cases and controls are not different with respect to socioeconomic status (chi square = 1.30, df = 1 and 0.26). Or we could report that the odds ratio (0.64) is not different from 1. We get this from the fact that the 95% confidence interval for the odds ratio, namely (0.27, 1.46), contains 1. We have that 66% of the controls and 76% of the cases are in the low socioeconomic category.. tab lung socio,row exp chi2 Key frequency expected frequency row percentage Lung cancer - patient/co Socioeconomic status ntrol Low High Total Control Patient Total Pearson chi2(1) = Pr = 0.255

6 . cc lung socio Proportion Exposed Unexposed Total Exposed Cases Controls Total Point estimate [95% Conf. Interval] Odds ratio (exact) Prev. frac. ex (exact) Prev. frac. pop chi2(1) = 1.30 Pr>chi2 = Step 6 is a teaching moment and not something I expected you to do. Suppose we thought that the difference of 66% of the controls in the low socioeconomic category versus 76% seemed like am important difference to us. Our sample is not particularly big so we decided to see exactly what size sample we would need to detect a difference of 10%. I first considered equal sample sizes and then because there were twice as many controls as cases, I looked at a 2 to 1 ratio. Below that we would have needed a sample size of 568 (ratio of 1) or 921 (ratio of 2 to 1).. sampsi Estimated sample size for two-sample comparison of proportions Test Ho: p1 = p2, where p1 is the proportion in population 1 and p2 is the proportion in population 2 Assumptions: alpha = (two-sided) power = p1 = p2 = n2/n1 = 1.00 Estimated required sample sizes: n1 = 534 n2 = 534

7 . sampsi ,r(2) p(0.80) Estimated sample size for two-sample comparison of proportions Test Ho: p1 = p2, where p1 is the proportion in population 1 and p2 is the proportion in population 2 Assumptions: alpha = (two-sided) power = p1 = p2 = n2/n1 = 2.00 Estimated required sample sizes: n1 = 307 n2 = 614