Question

An experimenter wishes to obtain a CI for the difference between true average breaking strength for cables manufactured by company I and by company II. Suppose breaking strength is normally distributed for both types of cable with s1 5 30 psi and s2 5 20 psi.

1. If costs dictate that the sample size for the type I cable should be three times the sample size for the type II cable, how many observations are required if the 99% CI is to be no wider than 20 psi?
2. Suppose a total of 400 observations is to be made. How many of the observations should be made on type I cable samples if the width of the resulting interval is to be a minimum?

Short answer

a.) n=47 ; m=141; b.) n=160, m=240.

Step-by-step solution

To Find the 20 psi

\((a):\)

For m, n large enough, confidence interval for \({\mu _1} - {\mu _2}\)with confidence level of approximately\(100(1 - \alpha )\% \)is

\(\bar x - \bar y \pm {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \)

where + and - give the adequate upper/lower limit of the interval. By replacing\({z_{\alpha /2}}\)by\({z_\alpha }\)and\( \pm \)with only + and - an upper/lower bound is obtained.

Assumption of normality indicates that the mentioned\(Cl\)needs to be used. The width of the interval is

\({\rm{\backslash begin}}alignedw{\rm{ \& }} = \bar x - \bar y + {z_{\alpha /2}} \cdot \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} - \left( {\bar x - \bar y - {z_{\alpha /2}} \cdot \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} } \right){\rm{n\& }} = 2 \cdot {z_{\alpha /2}} \cdot \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} .{\rm{n\backslash end}}aligned\)\[\]\(\begin{array}{l}w = \bar x - \bar y + {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} - \left( {\bar x - \bar y - {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} } \right)\\ = 2 \times {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} .\end{array}\)

For the 99 % confidence interval\(\alpha = 0.01\); thus \({z_{\alpha /2}} = {z_{0.01/2}} = 2.58\)

Find the t value 

The required width is 20 ; hence

\[\begin{array}{l}w = 2 \times {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \\20 = 2 \times 2.58 \times \sqrt {\frac{{3{0^2}}}{{3n}} + \frac{{2{0^2}}}{n}} \\3.876 = \sqrt {\frac{{3{0^2}}}{{3n}} + \frac{{2{0^2}}}{n}} \\\frac{{900 + 400 \times 3}}{{3n}} = 15.023\\3n = 139.786\\n = 46.6\end{array}\]

which means that the necessary sample size is

n=47 .

And for the I sample is m=3 n

m=141 .

Find the m value

(b):

It is required to minimize width

\(w = 2 \times {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \)\(w = 2 \times {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \)

However, since the total of 400 observations is to be made, one sample size is n and the other is 400-n (which makes it total of 400-n+n=400 ).

To minimize width w, look at w as a function of sample size n. To minimize function w, the following needs to minimized

\(f(n) = \frac{{s_1^2}}{{400 - n}} + \frac{{s_2^2}}{n}\)\(f(n) = \frac{{s_1^2}}{{400 - n}} + \frac{{s_2^2}}{n}\)

because m=400-n and width

\(w = 2 \times {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{{400 - n}} + \frac{{s_2^2}}{n}} \)\(w = 2 \times {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{{400 - n}} + \frac{{s_2^2}}{n}} \)

Taking derivative of f(n) in respect of n and equating to 0 the width will be minimized; thus

\(\) \(\begin{array}{l}{f^}(n) = {\left( {\frac{{900}}{{400 - n}} + \frac{{400}}{n}} \right)^}\\ = 900 \times \frac{1}{{{{(400 - n)}^2}}} - 400 \times \frac{1}{{{n^2}}}\end{array}\)

This yields

\(\begin{array}{l}9{n^2} = 4 \times {(400 - n)^2}\\9{n^2} = 4 \times \left( {40{0^2} - 2 \times 400 \times n + {n^2}} \right)\end{array}\)\(\begin{array}{l}9{n^2} = 4 \times {(400 - n)^2}\\9{n^2} = 4 \times \left( {40{0^2} - 2 \times 400 \times n + {n^2}} \right)\end{array}\)

or equally

\(m = 400 - {n_1} = 400 - 160 = 240\)\(5{n^2} + 3200n - 640,000 = 0\)

This is quadratic equation with two solutions

\(\begin{array}{l}{n_{12}} = \frac{{ - 3200 \pm \sqrt {320{0^2} - 4 \times 5 \times ( - 640,000)} }}{{2 \times 5}}\\{n_{12}} = \frac{{ - 3200 \pm 4800}}{{10}}\\{n_1} = \frac{{ - 3200 + 4800}}{{10}} = 160\\{n_2} = \frac{{ - 3200 - 4800}}{{10}} = - 800\end{array}\)

Choose \({n_1}\)because sample size can not be negative. The sample size m is therefore

\(m = 400 - {n_1} = 400 - 160 = 240\)

Final proof

a.) n=47 ; m=141; b.) n=160, m=240.