Question

1.      A quality control engineer is interested in the mean length of sheet insulation being cut automatically by machine. The desired mean length of the insulation is 12 feet. It is known that the standard deviation in the cutting length is 0.15 feet. A sample of 60 cut sheets yields a mean length of 12.14 feet. This sample will be used to obtain a 99% confidence interval for the mean length cut by machine. Calculate the confidence interval for the population mean length of the insulation. (2)

 

2.      The head of a computer science department is interested in estimating the proportion of students entering the department who will choose the new computer engineering option. Suppose there is no information about the proportion of students who might choose the option. What size sample should the department head take if she wants to be 95% confident that the estimate is within 0.05 of the true proportion? (1)

Answer 1

1. $\bar{x} = 12.14$

σ = 0.15

n = 60

α = 1 - 0.99 = 0.01

Standard error of mean $= \frac{σ}{\sqrt{n}} = \frac{0.15}{\sqrt{60}} = 0.0193$

The margin of error is:

$ME = \frac{σ}{\sqrt{n}} \times Z_{α/2} \\ = \frac{0.15}{\sqrt{60}} \times Z_{0.005} \\ = 0.0193 \times 2.575829 \\ = 0.0497$

A 99 % confidence interval for the mean length cut by machine:

$= \bar{x} ± ME \\ = 12.14 ± 0.0497 \\ = (12.0903, 12.1897)$

2. Margin of error E = 0.10

Significance level α = 0.05

The critical value

$Z_{α/2}=Z_{0.05/2}=1.96$

The Excel function is

=NORMSINV(0.05/2)

Since, the estimate of the population proportion is unknown.

Therefore, assume that

$\hat{p}=0.5$

Therefore, the required sample size is

$n = \hat{p}(1- \hat{p}) (\frac{Z_{α/2}}{E})^2 \\ = 0.5(1 -0.5)(\frac{1.96}{0.05})^2 \\ = 384$