Math  /  Data & Statistics

QuestionFast computer: Two microprocessors are compared on a sample of 6 benchmark codes to determine whether there is a difference in speed. The times (in seconds) used by each processor on each code are as follows: \begin{tabular}{ccccccc} \hline & \multicolumn{6}{c}{ Code } \\ \hline & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline Processor A & 28.9 & 17.1 & 21.8 & 17.6 & 20.5 & 26.4 \\ \hline Processor B & 22.4 & 18.1 & 28.9 & 28.4 & 24.7 & 27.5 \\ \hline \end{tabular} Send data to Excel
Part: 0/20 / 2
Part 1 of 2 (a) Find a 98%98 \% confidence interval for the difference between the mean speeds. Let dd represent the speed of processor A minus the speed of processor B . Use the TI-84 Plus calculator. Round the answers to two decimal places.
A 98\% confidence interval for the difference between the mean speeds is \square <μd<<\mu_{d}< \square .

Studdy Solution
Calculate the confidence interval for the mean difference μd\mu_d.
Margin of error=t×sdn=3.365×5.9568.18\text{Margin of error} = t^* \times \frac{s_d}{\sqrt{n}} = 3.365 \times \frac{5.95}{\sqrt{6}} \approx 8.18
The confidence interval is:
dˉMargin of error<μd<dˉ+Margin of error\bar{d} - \text{Margin of error} < \mu_d < \bar{d} + \text{Margin of error}
2.958.18<μd<2.95+8.18-2.95 - 8.18 < \mu_d < -2.95 + 8.18
11.13<μd<5.23-11.13 < \mu_d < 5.23
The 98% confidence interval for the difference between the mean speeds is:
11.13<μd<5.23 \boxed{-11.13 < \mu_d < 5.23}

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