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38. A recent article in The Wall Street Journal reported that the 30-year mortgage rate is now less than 6 percent. A sample of eight small banks in the Midwest revealed the following 30-year rates (in percent): 4.8 5.3 6.5 4.8 6.1 5.8 6.2 5.6 At the .01 significance level, can we conclude that the 30-year mortgage rate for small banks is less than 6 percent? Estimate the p-value. Solution: Mean of the sample = X = 5.6375 Standard deviation of the sample = σ = 0.63457 Null Hypothesis: H0: 0 = 6 vs. Ha:a < 6 Rejection region can be defined as z < -z Here significance level is 0.01, So, z0.01 = - 2.998 [critical value](Using Statistical Ratio Calculator From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating z with 0.01 significance) Now, z = (X - μ) / σx Where X is a normal random variable, μ is the mean, and σ is the standard deviation. x n Where n is the sample size. Calculating t-test statistics: z = (5.6375-6)/(0.63457/√ 8) = -1.616 Calculating p-value: Degree of freedom = DF = 8-1 = 7 P(z8 < -1.616) = 0.075064 (Using http://www.danielsoper.com/statcalc/calc08.aspx with t = -1.616 and DF = 7) The p-value is 7.5% which is greater than significance level of 1%. Hence, we fail to reject the null hypothesis but we can’t conclude that the rates are less than 6%. 32. Dole Pineapple, Inc., is concerned that the 16-ounce can of sliced pineapple is being overfilled. Assume the standard deviation of the process is .03 ounces. The quality control department took a random sample of 50 cans and found that the arithmetic mean weight was 16.05 ounces. At the 5 percent level of significance, can we conclude that the mean weight is greater than 16 ounces? Determine the p-value. Solution: Null Hypothesis: H0: 0 = 16 vs. Ha:a > 10 Rejection region can be defined as z > z Here significance level is 0.05, So, z0.05 = 1.645 [critical value](Using Statistical Ratio Calculator From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating z with 0.05 significance) Now, z = (X - μ) / σx Where X is a normal random variable, μ is the mean, and σ is the standard deviation. x n Where n is the sample size. Calculating t-test statistics: z = (16.05-16)/(0.03/√ 50) = 11.785 Calculating p-value: Degree of freedom = DF = 50-1 = 49 P(z49 > 11.785) = 0.000000 (Using http://www.danielsoper.com/statcalc/calc08.aspx with t = 11.785and DF = 49) The p-value of 0.00% is less than significant level of 5%. Hence, we can reject the null hypothesis and conclude that the mean weight is greater than 16 ounces.