SOLUTION: A TV special said there is a 10 per cent incidence of sexually transmitted disease (STD) among all U.S. teens. A reporter for the Rockville Bugle, looking for a story, surveyed 260

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: A TV special said there is a 10 per cent incidence of sexually transmitted disease (STD) among all U.S. teens. A reporter for the Rockville Bugle, looking for a story, surveyed 260      Log On

Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

   


Question 175855: A TV special said there is a 10 per cent incidence of sexually transmitted disease (STD) among all U.S. teens. A reporter for the Rockville Bugle, looking for a story, surveyed 260 randomly chosen teenagers, and found that 39 had been to a clinic for STD treatment. Use a right-tail test at α = .05.

The test statistic is
The critical value is
The p-value is

Answer by stanbon(48568) About Me  (Show Source):
You can put this solution on YOUR website!
A TV special said there is a 10 per cent incidence of sexually transmitted disease (STD) among all U.S. teens. A reporter for the Rockville Bugle, looking for a story, surveyed 260 randomly chosen teenagers, and found that 39 had been to a clinic for STD treatment. Use a right-tail test at α = .05.

The test statistic is
z(39/206) = ((39/206)-(0.10))/sqrt[0.1*0.9/260]= 4.2733
--------------------------------
The critical value is 1.645
--------------------------------
The p-value is
P(z>4.2733) = 0.000009637....
======================================
Cheers,
Stan H.