A study of 35 golfers showed that thier average score on aparticular course was

Question

A study of 35 golfers showed that thier average score on aparticular course was 92. The standard deviation of the populationis 5. What is the best point estimate of the mean? What is 95% confidence interval of the mean score for allgolfers? What is the 95%, if the sample of 60 golfers is used insteadof 35? Which is smaller? Show work, top rating. A study of 35 golfers showed that thier average score on aparticular course was 92. The standard deviation of the populationis 5. What is the best point estimate of the mean? What is 95% confidence interval of the mean score for allgolfers? What is the 95%, if the sample of 60 golfers is used insteadof 35? Which is smaller? Show work, top rating.

Explanation / Answer

average, = 92 standard deviation , = 5 standard error, e = / n = 5/35 =0.85 at 95% confidence for a two tail distribution, z = 1.96 95% confidence interval of the mean score for all golfers = ± (z*e) = 92 ± (1.96*0.85) = ( 90.334, 93.666) if the sample of 60 golfers is used , standard error, e = / n = 5/60 = 0.65 if the sample of 60 golfers is used , standard error, e = / n = 5/60 = 0.65 95% confidence interval of the mean score for all golfers= ± (z*e) = 92 ± (1.96*0.65) = ( 90.726, 93.274) the confidence interval for sample size = 60 issmaller

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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