Suppose a sample of 49 paired differences that have been randomly selected from a normally distributed population of paired differences yields a sample mean of d = 5 and a sample standard deviation of 5d: 7. (a) Calculate a 95 percent confidence interval for pd = #1 #2. (Round your answers to 2 decimal places.) Condence interval = [ . 1: (b) Test the null hypothesis H0: pd: 0 versus the alternative hypothesis Ha; pd: 0 by setting 0 equal to .10, .05, .01, and .001. How much evidence is there that yd differs from 0? t = Reject H0 at 0 equal to I , evidence that p1 differs from p2. (c) The p-value for testing H0: yd: 3 versus Ha: yd > 3 equals 0.0256. Use the p-value to test these hypotheses with 0: equal to .10, .05, .01, and .001. How much evidence is there that pd exceeds 3? What does this say about the size ofthe difference between 1.11 and 1.12? (Round your p-value answer to 4 decimal places.) p : Reject H0 at 0 equal to v , evidence that 111 and [12 differ by more than 3. A loan officer compares the interest rates for 48-month fixed-rate auto loans and 48-month variable-rate auto loans. Two independent, random samples of auto loan rates are selected. A sample of five 48-month variable-rate auto loans had the following loan rates: 2. 68 3. 078 2. 8728 3. 248 3. 158 while a sample of five 48-month fixed-rate auto loans had loan rates as follows: 4. 0328 3. 858 4. 3858 3. 758 4. 168 Click here for the Excel Data File FIGURE 11.7 JMP Output of Testing the Equality of Mean Loan Rates for Variable and Fixed 48-Month Auto Loans (for Exercise 11.12) Means and Std Deviations Level Number Mean Std Dev Fixed 4.03540 0.251785 Variable 2.98640 0.255176 t Test Variable-Fixed Assuming equal variances Difference -1.0490 t Ratio -6.54321 Std Err Dif 0.1603 DF 8 Upper CL Dif -0.6793 Prob > Itl 0.0002* Lower CL Dif -1.4187 Prob > t 0.9999 Confidence 0.95 Prob 0'22 with a = .05. What do you conclude? (Round your answers to F to the nearest whole number and F925 to 2 decimal places.) F = F0.05 = f H0: 0'21 5 0'2 An article in Fortune magazine reported on the rapid rise of fees and expenses charged by mutual funds. Assuming that stock fund expenses and municipal bond fund expenses are each approximately normally distributed, suppose a random sample of12 stock funds gives a mean annual expense of1.58 percent with a standard deviation of 0.48 percent, and an independent random sample of 12 municipal bond funds gives a mean annual expense of 0.97 percent with a standard deviation of 0.24 percent. Let ,ul be the mean annual expense for stock funds, and let [.12 be the mean annual expense for municipal bond funds. Do parts a, b, and c by using the equal variances procedure. (a) Set up the null and alternative hypotheses needed to attempt to establish that the mean annual expense for stock funds is larger than the mean annual expense for municipal bond funds. Test these hypotheses at the 0.05 level of signicance. (Round your 5P2 answer to 4 decimal places and t-value to 2 decimal places.) H0: p1 p2 versus Ha: u1 u2 32p t= 3.94 H0 with a = 0.05 (b) Set up the null and alternative hypotheses needed to attempt to establish that the mean annual expense for stock funds exceeds the mean annual expense for municipal bond funds by more than 0.5 percent. Test these hypotheses at the 0.05 level of signicance. (Round your t-value to 2 decimal places and other answers to 1 decimal place.) HO: u1 u2 versus Ha: u1 u2 [ t: H0 with u = 0.05 (c) Calculate a 95 percent confidence interval for the difference between the mean annual expenses for stock funds and municipal bond funds. Can we be 95 percent confident that the mean annual expense for stock funds exceeds that for municipal bond funds by more than .5 percent? (Round your answers to 3 decimal places.) The interval = [ the interval is 0.5.Suppose we have taken independent, random samples of sizes n1 = 8 and n2 = 7 from two normally distributed populations having means u1 and #2, and suppose we obtain x1 = 244 , X2 = 201 , s1 = 6, s2 = 4. Use critical values to test the null hypothesis Ho : #1 - M2 33 by setting a equal to .10, .05, .01 and .001. Using the equal variance procedure, how much evidence is there that the difference between 1 and /2 exceeds 33? (Round your answer to 3 decimal places.) t = HO at a = 0.1, 0.05, and, 0.01, evidence