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# Two-proportion z-test

last edited by 5 years, 8 months ago

A Two-proportion z-test: Test the null hypothesis $H_{0}:&space;p_{1}&space;-&space;p_{2}&space;=&space;0$ by comparing the statistic     $z&space;=&space;\frac{\hat{p}_{1}&space;-&space;\hat{p}_{2}}{SE_{\text{pooled}}(\hat{p}_{1}&space;-&space;\hat{p}_{2})}$

This is very useful when comparing two different proportions. Using the two-proportion z-test, we can find the difference in two different proportions. The two-proportion z-test shows the significance between the two proportions.

For example: Is Prozac more effective at treating Anorexia than a placebo? It compares these two proportions by finding out how many women were helped by each of the two drugs and giving us a statistic to show which drug worked in comparison to each other. Using the two-proportional z-test we can find the difference and see if the Prozac is more helpful than a placebo.  Our sample has 93 total women, 49 who received prozac with 35 who were deemed healthy, and 44 who received the placebo with 32 who were deemed healthy.

We are testing the hypothesis.

The population at interest is all women with anorexia. The sample is women who are being treated for anorexia.

The null hypothesis is that prozac will not have a difference in the treatment of women's anorexia compared to a placebo. The alternate hypothesis is that the prozac will treat the anorexia significantly better than the placebo.

We will use a normal model for our sampling distribution because we are sampling proportions.

Now check the relevant conditions for a normal model.

1) We assume that our data is random and representative.

2) It passes the 10% condition because our sample is only 93 women we are sure that is far below the total number of women with anorexia.

3) Now we need to test the success failure condition using the pooling method.

When testing a two-proportional z-test, we use a method called pooling. This is combining the counts together to get an overall proportion. We pool data to get better estimates when we use data from different sources. Because we want to believe that they really came from the same underlying population.

Formula for pooling proportions: $\hat{p}&space;=&space;\frac{x_{1}&space;+&space;x_{2}}{n_{1}&space;+&space;n{2}}$

When the conditions are met we can continue with the hypothesis test.

We put this combined value into the formula for standard error, substituting it for both sample proportions.

Now we find the test statistics.

Calculate the P-Value. We used spss to calculate the p-value which is .889.(See Chi-Squared table below)

With a P-Value this high we fail to reject the Null hypothesis. There is insufficient evidence to support that prozac has a significance in treating anorexia over a placebo. There is not a high possibility for a type II error but there is a very slim chance.

As you can see from this clustered bar chart from SPSS that there is not a significant difference between the placebo and Prozac.

Generating a crosstab output for a two-proportional z-test in spss

• Go to analyze, descriptive statistics, crosstabs
• Move your explanatory variable into the rows box and the response variable into the columns box.
• Click statistics then select Chi Squared and then click continue
• Click Cells then select rows in the percentages box and click continue (This makes things much easier to look at)
• Click okay
• The crosstabs output will appear.
• Find the Chi-squared test table. Look at the persons Chi-squared row and asymp. sig (2 sided) column. This is where you will find your P-value