1) Test the association between smoking and esophageal cancer using SPSS.

Here the null hypothesis is

Ho: There is no association between smoking and esophageal cancer.

and the alternative hypothesis is

Ha: There is no association between smoking and esophageal cancer.

The Expected counts are calculated in the following table.

Since the p-value corresponding to the Chi-Square test is 0.000 < 0.05, there exist significant association between smoking and esophageal cancer.

For doing this Analysis in SPSS,

Form the main menu select

Analyze -> Descriptive Statistics -> Crosstabs

Then scoot Tobacco consumption into the Row box and Esophageal cancer into the Column box and then click Continue, OK.

In order to compute the expected frequencies, do the above steps and click cells tab and tick the option Expected.

In order to obtain the Chi-square test statistic, do the above steps and click Statistics tab and tick the option Chi-square.

Test the association between the dichotomized smoking variable and esophageal cancer using SPSS.

Here the null hypothesis is

Ho: There is no association between smoking and esophageal cancer.

and the alternative hypothesis is

Ha: There is no association between smoking and esophageal cancer.

The Expected counts are calculated in the following table.

Since the p-value corresponding to the Chi-Square test is 0.000 < 0.05, there exist significant association between the dichotomized smoking variable and esophageal cancer.

2) Odds ratio = OR = (a*d)/(b*c), where a and b are the items in the first row and c and d are the items in the second row.

Here, a = 64, b = 150, c=136 and d = 625.

Therefore, OR = (64*625)/(150*136) = 1.9608

Confidence Interval for the OR is Exp[Ln(OR) ± Z-score*SE(OR)], where

SE(OR) = the standard error of the odds ratio = Sqrt[(1/a)+(1/b)+(1/c)+(1/d)]

= Sqrt[(1/64)+(1/150)+(1/136)+(1/625)] = 0.1768

and Z-Score corresponding to 95% confidence level = 1.96

Therefore, the OR 95% confidence interval for the association between the dichotomized smoking variable and esophageal cancer is given by,

(Exp[Ln(1.9608) - 1.96*0.1768],Exp[Ln(1.9608) - 1.96*0.1768])

= (Exp(0.3269), Exp(1.0198))

= (1.3867, 2.7726)