Posted by **Anonymous** on Friday, March 2, 2012 at 4:58pm.

A random sample of 80 youths and a second random sample of 120 adults showed that 18 of the youths and 10 of the adults had been ticketed for careless driving. Use a 1% level of significance to test the claim that youth have a higher proportion of careless drivers than adults do. state the null and alternative hypothesis

- statistics -
**MathGuru**, Friday, March 2, 2012 at 6:35pm
You will need to use a formula for a binomial proportion two-sample z-test.

Proportion of adults = 10/120 = .083

Proportion of youths = 18/80 = .225

Use these proportions in the formula.

Null:

Ho: pY = pA (A = adults; Y = youths)

Alternative:

Ha: pY > pA

This will be a one-tailed test at .01 level of significance. (Use a z-table to determine the critical or cutoff value to reject the null.) If the test statistic calculated from the formula exceeds the critical value from the table, reject the null and conclude pY > pA. If the test statistic does not exceed the critical value from the table, do not reject the null (there is no difference).

I hope this brief explanation will help get you started.

- statistics -
**carlos**, Friday, March 2, 2012 at 7:12pm
(b) How many would you expect to be taller than 160 cm?

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