Pulse Rates of Identical Twins A researcher wanted to compare the pulse rates of identical twins to see whether there was any difference. Eleven sets of twins were randomly selected. The rates are given in the table as number of beats per minute. Find the $98 \%$ confidence interval for the difference of the average pulse rates of twins. Assume the variables are normally distributed. Let $\mu_{A}$ be the average of twin $\mathrm{A}$ and $\mu_{D}=\mu_{A}-\mu_{B}$. Round the answer to at least one decimal place.
\begin{tabular}{c|ccccccccccc}
Twin A & 85 & 92 & 88 & 93 & 78 & 87 & 76 & 91 & 83 & 86 & 90 \\
\hline Twin B & 84 & 80 & 93 & 86 & 95 & 83 & 78 & 79 & 83 & 89 & 86
\end{tabular}

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\[
\square< \mu_{D}< \square \quad x \quad 5
\]

Step 1 ：Given the pulse rates of 11 sets of twins, we first calculate the differences between each pair of twins' pulse rates. The differences are: 1, 12, -5, 7, -17, 4, -2, 12, 0, -3, 4.

Step 2 ：Next, we calculate the mean of these differences, which is approximately \(1.18\).

Step 3 ：We also calculate the standard deviation of these differences, which is approximately \(8.26\).

Step 4 ：Given that the number of pairs of twins is 11, the degrees of freedom is \(11 - 1 = 10\).

Step 5 ：Given a confidence level of 98%, the alpha level is \(1 - 0.98 = 0.02\).

Step 6 ：Using a t-distribution table or calculator, we find the critical t-value for a two-tailed test with 10 degrees of freedom and an alpha level of 0.02 to be approximately \(2.76\).

Step 7 ：We then calculate the margin of error by multiplying the critical t-value by the standard deviation and dividing by the square root of the number of pairs of twins. The margin of error is approximately \(6.88\).

Step 8 ：Finally, we calculate the confidence interval by subtracting and adding the margin of error from the mean difference. The 98% confidence interval for the difference of the average pulse rates of twins is \(\boxed{(-5.7, 8.1)}\) beats per minute.