Posted by **Mary** on Tuesday, March 30, 2010 at 12:28am.

Inferences about Two Proportions technique (Chapter 9):

There is a claim that proportion of successes increased in 2009 compare to 2008.

For data given in Table 1

Table 1

Number of cases Number of successes

2008 400 180

2009 500 235

Calculate pooled sample proportion and z-test statistics

Use a significance level of á = 0.05 to make decision if there is a considerable change of proportion of successes.

- STATS -
**MathGuru**, Tuesday, March 30, 2010 at 7:41pm
This would fit a one-tailed proportional z-test (using proportions). The test would be one-tailed because of the alternate hypothesis (p2009 > p2008).

Here is one formula:

z = (p2009 - p2008)/√[pq(1/n1 + 1/n2)]

...where 'n' is the sample sizes, 'p' is (x1 + x2)/(n1 + n2), and 'q' is 1-p.

I'll get you started:

p = (180 + 235)/(400 + 500) = ? -->once you have the fraction, convert to a decimal (decimals are easier to use in the formula).

q = 1 - p

p2008 = 180/400

p2009 = 235/500

Convert all fractions to decimals. Plug those decimal values into the formula and find z. Once you have this value, compare to the critical value from a z-table using 0.05 significance for a one-tailed test.

If the test statistic exceeds the critical value from the table, reject the null and accept the alternate hypothesis (p2009 < p2008). If the test statistic does not exceed the critical value from the table, then you cannot reject the null and you cannot conclude a difference.

I hope this will help.

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