**Pulse Rates of Identical Twins**A researcher wanted to compare the pulse rates of identical twins to see whether there was any difference. Twelve sets of twins were randomly selected. The rates are given in the table as number of beats per minute. Find the 90% 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 A and \( \mu_D = \mu_A - \mu_B \). Round the answer to at least one decimal place.| Twin A | 85 | 83 | 84 | 93 | 86 | 78 | 76 | 92 | 91 | 90 | 88 | 87 ||--------|----|----|----|----|----|----|----|----|----|----|----|----|| Twin B | 89 | 93 | 80 | 86 | 84 | 78 | 83 | 93 | 79 | 95 | 86 | 83 |A button is available to send data to Excel for further analysis.[Interactive features for entering and checking the confidence interval calculations are displayed below the table.]

Followings are the Explanation of the question Use the Excel and calculate the Mean and Stanard deviation for the Difference d= Twin A - Twin BConstruct 90% confidence for the Differences of the averages μd The sample mean for difference Xd = 0.333 Sample standard deviation for difference sd = 6.198 The size of the sample is n = 12 The required confidence level is 90% Find the Level of significance α 100(1-alpha) %= 90% 100(1-alpha) = 100 (1-alpha) = 90/100 (1-alpha) = 0.90 alpha = 1-0.90 alpha = 0.10 Use the t-distribution table and find the t -critical value Followings are the Steps to find the critical value 1) Use the t-table 2) use the formula for critical value 3) calculate the degrees of freedom (DF) = n-1= 12-1=11 4) Find the 1−α/2 value = 0.95 5) See the t table and in row see the 11 number and in column see the 0.95 number 6) there interaction value is a critical value Based on the provided information, the critical t-value for α=0.10 is tc=t(0.10⁄2,n-1)=1.7959 tcritical=1.7959---------------using t table The 90% confidence for the Differences of the averages μd is computed using the following expression CI=[Xd-t(α⁄2)*sd/√n],[Xd+t(α⁄2)*sd/√n]CI=[0.333-1.7959*6.198/√12],[0.333+1.7959*6.198/√12] CI=[-2.8,3.5] Where Margin of Error=t(α⁄2)*s/√n=3.21Lower Limit=([X)¯–t(α⁄2)*s/√n]=-2.8Upper Limit=([X)¯+t(α⁄2)*s/√n]=3.5 Therefore, based on the information provided, the 90 % confidence for the Difference average μd is [ -2.8< μd< 3.5].