To test Upper H 0 : pequals 0.50 versus Upper H 1 : pnot equals 0.50, a simple r

Question

To test Upper H 0 : pequals 0.50 versus Upper H 1 : pnot equals 0.50, a simple random sample of nequals 500 individuals is obtained and xequals 245 successes are observed.

(a) What does it mean to make a Type II error for this test?

(b) If the researcher decides to test this hypothesis at the alpha equals0.10 level of significance, compute the probability of making a Type II error, beta , if the true population proportion is 0.55 . What is the power of the test?

(c) Redo part (b) if the true population proportion is 0.52 .

If you could exsplain how to do it with a TI-89

Explanation / Answer

H0 : p = 0.50

Ha : p 0.50

Sample size = 500

success = 245

p^= 245/500 = 0.49

(a) Here type II error means that is incorrectly retaining a false null hypothesis. So in this case, we will accept that true proportion of successes are 0.50 even if it is untrue.

(b) Here significane level = 0.10

standard error of proportion = sqrt (p0 * (1-p0)/N) = sqrt(0.5 * 0.5/ 500) = 0.0224

so, the confidence interval = p^ +- Z90% sqrt (p0 * (1-p0)/N)

= 0.49 +- 1.645 * sqrt (0.5 * 0.5/ 500)

= 0.49 +- 0.0368

= (0.4532, 0.5268)

If true population mean is 0.55 then what is the probability that we will fail to reject it. We will fail to reject it when the sample proportion is under the limit of confidence interval.

so Pr( p^ < 0.5268, 0.55, 0.0224)

Z = (0.5268 - 0.55)/ 0.0224 = -1.04

so Pr( Z < -1.04) = 0.1492

Power of the test = 1 - 0.1492 = 0.8508

(c) If true population proportion = 0.52

standard error of proportion = sqrt (p0 * (1-p0)/N) = sqrt(0.5 * 0.5/ 500) = 0.0224

so, the confidence interval = p^ +- Z90% sqrt (p0 * (1-p0)/N)

= 0.49 +- 1.645 * sqrt (0.5 * 0.5/ 500)

= 0.49 +- 0.0368

= (0.4532, 0.5268)

If true population mean is 0.55 then what is the probability that we will fail to reject it. We will fail to reject it when the sample proportion is under the limit of confidence interval.

so Pr( p^ < 0.5268, 0.52, 0.0224)

Z = (0.5268 - 0.52)/ 0.0224 = 0.30

so Pr( Z < 0.30) = 0.6179

Power of the test = 1 - 0.6179 = 0.3821

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.