# statistics

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From a random sample of 500 males interviewed, 125 indicated that they watch professional football on Monday night television. Does this evidence indicate that more than 20% of the male TV viewers watch professional football on Monday evenings? Use the 0.01 level of significance, and find the p-value.

• statistics -

You can try a proportional one-sample z-test for this one since this problem is using proportions.

Here's a few hints to get you started:

Null hypothesis:
Ho: p = .20 -->meaning: population proportion is equal to .20

Alternative hypothesis:
Ha: p > .20 -->meaning: population proportion is greater than .20

Using a formula for a proportional one-sample z-test with your data included, we have:
z = .25 - .20 -->test value (125/500 is .25) minus population value (.20) divided by
√[(.25)(.75)/500] --> .75 represents 1-.25 and 500 is sample size.

Finish the calculation. Remember if the null is not rejected, then you cannot conclude p > .20. If you need to find the p-value for the test statistic, check a z-table, then compare to .01 to determine whether or not to reject the null.

I hope this helps.

• Correction -

Using a formula for a proportional one-sample z-test with your data included, we have:
z = .25 - .20 -->test value (125/500 is .25) minus population value (.20) divided by
√[(.20)(.80)/500] --> .80 represents 1-.20 and 500 is sample size.

Sorry for any confusion.

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