Question

nick1337 · Accepted Answer

To find the 90% confidence interval for the difference in mean parts produced per hour by machines A and B, we can use the two-sample t-test formula. The formula is given as:

\bar{X_{1}} - \bar{X_{2}} \pm t_{\frac{α}{2}, n_{1} + n_{2} - 2} \times \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}

where

\bar{X_{1}}

and

\bar{X_{2}}

are the sample means for machines A and B,

s_{1}

and

s_{2}

are the sample standard deviations for machines A and B,

n_{1}

and

n_{2}

are the sample sizes for machines A and B, and

t_{\frac{α}{2}, n_{1} + n_{2} - 2}

is the critical value of the t-distribution with

n_{1} + n_{2} - 2

degrees of freedom and a significance level of

\frac{α}{2}

.
Substituting the given values, we get:

\bar{X_{1}} - \bar{X_{2}} \pm t_{\frac{0.10}{2}, 50 + 45 - 2} \times \sqrt{\frac{{4.6}^{2}}{50} + \frac{{6.8}^{2}}{45}}

Simplifying the expression, we have:

230 - 215 \pm t_{0.05, 93} \times \sqrt{0.0424}

To find the critical value

t_{0.05, 93}

, we can use a t-table or a calculator. Using a calculator, we get:

t_{0.05, 93} \approx 1.664

Substituting this value, we get:

15 \pm 1.664 \times 0.2057

Simplifying the expression, we get:

15 \pm 0.342

Therefore, the 90% confidence interval for the difference in mean parts produced per hour by machines A and B is:

(14.658, 15.342)

This means that we are 90% confident that the true difference in mean parts produced per hour by machines A and B lies between 14.658 and 15.342 units.

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