Question

A machine produces parts with lengths that are normally distributed with σ = 0.6. A sample of 19 parts has a mean length of 75.28.

(a) Give a point estimate for μ. (Give your answer correct to two decimal places.)

(b) Find the 98% confidence maximum error of estimate for μ. (Give your answer correct to three decimal places.)

(c) Find the 98% confidence interval for μ. (Give your answer correct to three decimal places.)

Lower Limit Upper Limit You may need to use the appropriate table in Appendix B to answer this question.

Answer 1

Given,

$\sigma$= 0.6

sample mean = $\bar{x}$ = 75.28

Sample size = n = 19

a )

Sample mean ( $\bar{x}$ ) is point estimate of population mean $\mu$

point estimate for μ = $\bar{x}$ = 75.28

b)

Maximum error of estimate for μ is ,

$ME = z_{\alpha /2}*\frac{\sigma }{\sqrt{n}}$

Where, $z_{\alpha /2}$ is critical value at given confidence level

Confidence level = 98% = 0.98

Significance level = $\alpha$ = 1 - 0.98 = 0.02 , $\alpha$/2 = 0.01

$z_{\alpha /2}$ = $z_{0.01}$ = 2.3264                    { Using Excel, = NORMSINV( 0.01 ) = 2.3264 }

So, 98% confidence maximum error of estimate for μ is,

$ME = 2.3264 * \frac{0.6}{\sqrt{19}}$

ME = 0.320228 $\approx$0.320

c)

98% confidence interval for μ is,

Lower Limit = $\bar{x}$ - ME = 75.28 - 0.320 = 74.960

Upper Limit = $\bar{x}$ + ME = 75.28 + 0.320 = 75.600