Question

10. The lowest and highest observations in a population are 19 and 63, respectively. What is the minimum sample size n required to estimate μ with 95% confidence if the desired margin of error is E = 2.6? What happens to n if you decide to estimate μ with 90% confidence?

(You may find it useful to reference the z table. Round intermediate calculations to at least 4 decimal places and "z" value to 3 decimal places. Round up your answers to the nearest whole number.)

Confidence Level	n
95%	? (31 WRONG ?)
90%	? (22 WRONG ?)

Please do not answer if you are not sure, THANK YOU! :)

Answer 1

10

Maximum = 63

Minimum = 19

Range = maximum - minimum = 63 - 19 = 44

Using Range rule of thumb

Population Standard deviation = Range / 4 = 44 / 4 = 11

Margin of error = E = 2.6

At 95% confidence,

Z_/2 = Z_0.025 = 1.96

sample size = n = [Z_/2* / E] ²

n = [1.96 * 11 / 2.6]²

n = 69

Sample size = n = 69

At 90% confidence,

Z_/2 = Z_0.05 = 1.645

sample size = n = [Z_/2* / E] ²

n = [1.645 * 11 / 2.6]²

n = 49

Sample size = n = 49