Suppose we wish to demonstrate that there is a difference between the proportions of wives and husbands who do laundry at home. From a random sample of 66 randomly selected wives, we observe 44 who do laundry at home. From a random sample of 46 husbands, we observe 18 who do laundry at home. Test the claim that the proportion of wives, p 1 , who do laundry at home is different from the proportion of husbands, p 2 , who do laundry at home. Use a 1% significance level.

Step-by-step explanation:

From this study:

The null hypothesis:

The altenative is:

This test is a two-tailed test.

However; we are told that the wives have 44 success out of 66, then the number of failures will be 22.

Then;

Similarly, the husbands have 18 success out of 46, then the number of failures will be 28

Then:

The pooled proportion

p = 0.55357

The estimated standard error S.E is:

=

= 0.0955

The Z test statitics can now be computed as:

Z = 2.88

Th p -value from the test statistics is:

p-value = 2P(Z > 2.88)

p- value = 2 P (1 - Z < 2.88)

p-value = 2 ( 1 - 0.998)

p-value = 2 ( 0.002)

p -alue = 0.004

Decision Rule:

Thus, at 0.01 significance level, we reject the null hypothesis because, p-value is less than that (i.e. significance level)

Conclusion:

We conclude that there is a significant  difference between the proportions.

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