The U.S. Census Bureau collects data on the ages of married people. Suppose that eight married couples are randomly selected and have the ages given in the following table. Determine the 98%confidence interval for the true mean difference between the ages of married males and married females.Let d=(age of husband)−(age of wife). Assume that the ages are normally distributed for the populations of both husbands and wives in the U.S.Husband5128392531355941Wife6026363326436450 Step 1 of 4: Find the mean of the paired differences, d‾. Round your answer to one decimal place.Step 2 of 4: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.Step 3 of 4: Find the standard deviation of the paired differences to be used in constructing the confidence interval. Round your answer to one decimal placeStep 4 of 4: Construct the 98% confidence interval. Round your answers to one decimal place

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Description: The U.S. Census Bureau collects data on the ages of married people. Suppose that eight married couples are randomly selected and have the ages given in the following table. Determine the 98%confidence interval for the true mean difference between the

Step 1 of 4: The difference is, Husband Wife Difference (d) 51 60 -9 28 26 2 39 36 3 25 33 -8 31 26 5 35 43 -8 59 64 -5 41 50 -9 The mean of the paired differences is, d¯=∑dn=-9+2+3-8+5-8-5-98=-298=-3.625≈-3.6 Thus, the mean of the paired differences is -3.6.Step 2 of 4: The degrees of freedom is, df=n-1=8-1=7 The degrees of freedom is 7. Computation of critical value: The critical value of t-distribution for 0.02 level of significance at 7 degrees of freedom can be obtained using the excel formula “=T.INV.2T(0.02,7)”. The critical value is 2.998. Thus, the critical value is 2.998.Step 3 of 4: The standard deviation of the paired difference is, sd=∑d-d¯2n-1=-9+3.6252+2+3.6252+3+3.6252+-8+3.6252+5+3.6252+-8+3.6252+-5+3.6252+-9+3.62528-1=247.87507=35.4107=5.95069≈6.0 Thus, the standard deviation of the paired difference is 6.Step 4 of 4: The 98% confidence interval is, CI=d¯±tα2sdn=-3.6±2.99868=-3.6±6.4=-3.6-6.4,-3.6+6.4=-10.0,2.8 Thus, the 98% confidence interval is (-10.0,2.8).

Answered: Step 1 of 4: Find the mean of the…

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The U.S. Census Bureau collects data on the ages of married people. Suppose that eight married couples are randomly selected and have the ages given in the following table. Determine the 98%confidence interval for the true mean difference between the ages of married males and married females.

Let d=(age of husband)−(age of wife). Assume that the ages are normally distributed for the populations of both husbands and wives in the U.S.

Husband	51	28	39	25	31	35	59	41
Wife	60	26	36	33	26	43	64	50

Step 1 of 4: Find the mean of the paired differences, d‾. Round your answer to one decimal place.

Step 2 of 4: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 3 of 4: Find the standard deviation of the paired differences to be used in constructing the confidence interval. Round your answer to one decimal place

Step 4 of 4: Construct the 98% confidence interval. Round your answers to one decimal place

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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