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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

STEP 1

1. The sample of benchmark codes is independent and identically distributed.
2. The differences in times are normally distributed.
3. The data provided is for paired samples, where each code is tested on both processors.
4. We are using a TI-84 Plus calculator to compute the confidence interval.

STEP 2

1. Calculate the differences between the times for each code.
2. Compute the mean and standard deviation of the differences.
3. Determine the critical value for a 98% confidence interval.
4. Calculate the confidence interval for the mean difference.

STEP 3

Calculate the differences di d_i for each code, where di=Processor A time−Processor B time d_i = \text{Processor A time} - \text{Processor B time} .
\begin{align*} d_1 &= 28.9 - 22.4 = 6.5 \\ d_2 &= 17.1 - 18.1 = -1.0 \\ d_3 &= 21.8 - 28.9 = -7.1 \\ d_4 &= 17.6 - 28.4 = -10.8 \\ d_5 &= 20.5 - 24.7 = -4.2 \\ d_6 &= 26.4 - 27.5 = -1.1 \\ \end{align*}

STEP 4

Compute the mean dˉ\bar{d} and standard deviation sds_d of the differences.
dˉ=6.5−1.0−7.1−10.8−4.2−1.16=−17.76=−2.95\bar{d} = \frac{6.5 - 1.0 - 7.1 - 10.8 - 4.2 - 1.1}{6} = \frac{-17.7}{6} = -2.95
Calculate the standard deviation sds_d:
sd=(6.5+2.95)2+(−1.0+2.95)2+(−7.1+2.95)2+(−10.8+2.95)2+(−4.2+2.95)2+(−1.1+2.95)26−1s_d = \sqrt{\frac{(6.5 + 2.95)^2 + (-1.0 + 2.95)^2 + (-7.1 + 2.95)^2 + (-10.8 + 2.95)^2 + (-4.2 + 2.95)^2 + (-1.1 + 2.95)^2}{6 - 1}}
sd=(9.45)2+(1.95)2+(−4.15)2+(−7.85)2+(−1.25)2+(1.85)25s_d = \sqrt{\frac{(9.45)^2 + (1.95)^2 + (-4.15)^2 + (-7.85)^2 + (-1.25)^2 + (1.85)^2}{5}}
sd=89.3025+3.8025+17.2225+61.6225+1.5625+3.42255s_d = \sqrt{\frac{89.3025 + 3.8025 + 17.2225 + 61.6225 + 1.5625 + 3.4225}{5}}
sd=176.9355=35.387≈5.95s_d = \sqrt{\frac{176.935}{5}} = \sqrt{35.387} \approx 5.95

STEP 5

Determine the critical value for a 98% confidence interval with n−1=5n-1 = 5 degrees of freedom using a t-distribution. Use the TI-84 Plus calculator to find t∗t^*.
For a 98% confidence interval, the critical value t∗t^* is approximately 3.365.

STEP 6

Calculate the confidence interval for the mean difference μd\mu_d.
Margin of error=t∗×sdn=3.365×5.956≈8.18\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:
dˉ−Margin of error<μd<dˉ+Margin of error\bar{d} - \text{Margin of error} < \mu_d < \bar{d} + \text{Margin of error}
−2.95−8.18<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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