The lowest and highest observations in a population are 17 and 57, respectively.

Question

The lowest and highest observations in a population are 17 and 57, respectively. What is the minimum sample size n required to estimate with 90% confidence if the desired margin of error is E = 2.8? What happens to n if you decide to estimate with 95% confidence? Use Table 1. (Round intermediate calculations to 4 decimal places and "z-value" to 3 decimal places. Round up your answers to the nearest whole number.)

  Confidence Level

n            

90%

95%

The lowest and highest observations in a population are 17 and 57, respectively. What is the minimum sample size n required to estimate with 90% confidence if the desired margin of error is E = 2.8? What happens to n if you decide to estimate with 95% confidence? Use Table 1. (Round intermediate calculations to 4 decimal places and "z-value" to 3 decimal places. Round up your answers to the nearest whole number.)

  Confidence Level

n            

90%

95%

Explanation / Answer

Since the value of standard deviation is not given, we consider Range as a measure of dispersion

Range = 57-17 = 40

for 90% confidence interval

n = (zR/E)2 = (1.654*40/2.8)2 = 558.3093 approx 558

For 95 % confidence interval

n = (zR/E)2 = (1.96*40/2.8)2 = 784

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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