Question

The diameter of a wheel from a certain plant may be considered to be normally distributed with variance 9 cm2.

1.1
If a sample of 30 wheels has a mean of 50 cm, give a 90% Confidence Interval for the mean diameter of the wheels from the factory.
Enter your answer as an interval [ a, b ] where a, b should be accurate to two decimal places.

1.2
If an interval of width (at most) 0.1cm is to contain the mean with a confidence of 95%, how large should the sample size be?
Give your answer as the smallest whole number which would ensure the interval is no wider than 0.1cm.

Answer 1

Solution :

Given that,

1.1) Point estimate = sample mean = $\bar x$ = 50 cm

variance = $\sigma$ ² = 9

Population standard deviation =   $\sigma$ = $\sqrt{}$$\sigma$² = $\sqrt{}$9 = 3

Sample size = n = 30

At 90% confidence level

$\alpha$ = 1 - 90%  

$\alpha$ = 1 - 0.90 =0.10

$\alpha$/2 = 0.05

Z$\alpha$/2 = Z_{0.05 = 1.645}

Margin of error = E = Z$\alpha$/2 * ( $\sigma$ /$\sqrt$n)

= 1.645 * ( 3 /  $\sqrt$30 )

= 0.90

At 90% confidence interval estimate of the population mean is,

$\bar x$  ± E

50 ± 0.90

( 49.10, 50.90)

1.2 ) Margin of error = E = 0.1

At 95% confidence level the z is,

$\alpha$ = 1 - 95%

$\alpha$ = 1 - 0.95 = 0.05

$\alpha$/2 = 0.025

Z_$\alpha$/2 = 1.96

sample size = n = [Z_$\alpha$/2* $\sigma$ / E] ²

n = [ 1.96 * 3 / 0.1]²

n = 3457.44

Sample size = n = 3457