Question

A random sample of nn measurements was selected from a population with standard deviation σ=19.9σ=19.9 and unknown mean μμ. Calculate a 9595 % confidence interval for μμ for each of the following situations:

(a) n=60, x¯¯¯=95.6n=60, x¯=95.6
≤μ≤≤μ≤

(b) n=80, x¯¯¯=95.6n=80, x¯=95.6
≤μ≤≤μ≤

(c) n=100, x¯¯¯=95.6n=100, x¯=95.6
≤μ≤≤μ≤

(d) In general, we can say that for the same confidence level, increasing the sample size  the margin of error (width) of the confidence interval. (Enter: ''DECREASES'', ''DOES NOT CHANGE'' or ''INCREASES'', without the quotes.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 95.6

Population standard deviation =    = 19.9

a) Sample size = n = 60

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_0.025 = 1.96

Margin of error = E = Z/2 * ( /n)

= 1.96 * ( 19.9 /  60 )

= 5.04

At 95% confidence interval estimate of the population mean is,

- E + E

95.6 - 5.04     95.6 + 5.04

( 90.56    100.64 )

b) Sample size = n = 80

Margin of error = E = Z/2 * ( /n)

= 1.96 * ( 19.9 /  80 )

= 4.36

At 95% confidence interval estimate of the population mean is,

- E + E

95.6 - 4.36     95.6 + 4.36

( 91.24     99.96 )

c) Sample size = n = 100

Margin of error = E = Z/2 * ( /n)

= 1.96 * ( 19.9 /  100 )

= 3.90

At 95% confidence interval estimate of the population mean is,

- E + E

95.6 - 3.90     95.6 + 3.90

( 91.70     99.50 )

d) In general, we can say that for the same confidence level, increasing the sample size  the margin of error (width) of the confidence interval DECREASES