Question

(a) How large a sample must be drawn so that a 98% confidence interval for μ will have a margin of error equal to 3.8? Round the critical value to no less than three decimal places. Round the sample size up to the nearest integer.

A sample size of __?is needed to be drawn in order to obtain a 98% confidence interval with a margin of error equal to 3.8.

(b) If the required confidence level were 99.5%, would the necessary sample size be larger or smaller?

(larger OR smaller) , because the confidence level is  (higher OR lower) .

Answer 1

Solution :

here standard deviation is not given iam assuming the standard deviation here.

Given,

a ) standard deviation = = 17.5 , Zc = Z0.05 = +/- 2.328

margin of error = E = 3.8

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z/2 = Z0.01 = 2.326

Sample size = n = ((Z/2 * ) / E)2

= ((2.326 * 17.5) / 3.8)2

= 114.74  

115

Sample size = 115

A sample size of _115_is needed to be drawn in order to obtain a 98% confidence interval with a margin of error equal to 3.8.

b)

(b) if the required confidence level were 99% would the neccessary sample size be larger or smaller

=> larger , because the confidence level is  Higher .

explanation:-

=> here confidence interval is 99.5 %

99.5 % confidence for z is 2.807

Zc = Z0.005 = 2.807

n = ((2.807 * 17.5) / 3.8)2

  = 167.10

167

Sample size be larger because the confidence level is Higher

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