Lind Chapter 9; Exercise 12

The American Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 16 people reveals the mean yearly consumption to be 60 pounds with a standard deviation of 20 pounds. a.What is the value of the population mean? What is the best estimate of this value? The value of population mean is unknown. The best estimate of this value is the sample mean of 60 pounds. b.Explain why we need to use the t distribution. What assumption do you need to make? According to Lind, et al. (2005), when population standard deviation is unknown, and the sample is smaller than 30, a t distribution should be used. We need to assume that the sample is from a normal population (pp. 291-293). c.For a 90 percent confidence interval, what is the value of t? For a 90 percent confidence interval, and df = 15, t= 1.753

d.Develop the 90 percent confidence interval for the population mean. Xbar = 60; s = 20; n = 16

Xbar ±t(s/√n) = 60 ± 1.753 (20/√16) = 60± 1.753 (5) = 60± 8.765 = [51.24, 68.77] e.Would it be reasonable to conclude that the population mean is 63 pounds? Yes because 63 pounds is among the confidence intervals.

Lind Chapter 9; Exercise 28

A processor of carrots cuts the green top off each carrot, washes the carrots, and inserts six to a package. Twenty packages are inserted in a box for shipment. To test the weight of the boxes, a few were checked. The mean weight was 20.4 pounds, the standard deviation 0.5 pounds. How many boxes must the processor sample to be 95 percent confident that the sample mean does not differ from the population mean by more than 0.2 pounds?

Xbar = 20.4; s = 0.5; E= 0.2; 95% confidence level

At 95% confidence level, z = 1.96

n = (zs/E)² = [(1.96)(0.5)/0.2]² = 24.01

24.01 boxes must the processor sample to be 95 percent confident that the same mean does not differ from the population mean by more than 0.2 pounds.

Lind Chapter 10; Exercise 6

The MacBurger restaurant chain claims that the waiting time of customers for service is normally distributed, with a mean of 3 minutes and a standard deviation of 1 minute. The quality-assurance department found in a sample of 50 customers at the Warren Road MacBurger that the mean waiting time was 2.75 minutes. At the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes? µ = 3; σ = 1; xbar = 2.75; n = 50; one-tailed test

At 0.05 significance level, critical value = -1.65 (found by 0.5-0.05 = 0.4500) H₀: The mean waiting time is greater than or equal to 3 minutes H₁: The mean waiting time is less than 3 minutes

Z = (xbar - µ)/σ/√n = (2.75-3)/1/√50 = -0.25/1/7.07 = -1.77 Since -1.77 falls in the rejection region, H₀ is rejected; or at the 0.05 significance level, we can conclude that the mean waiting time is less than 3 minutes. Lind Chapter 10; Exercise 18

The management of White Industries is considering a new method of assembling its golf cart. The present method requires 42.3 minutes, on the average, to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes. Using the .10 level of significance, can we conclude that the assembly time using the new method is faster? n = 24; df = 23; µ = 42.3; xbar = 40.6; s = 2.7; one-tailed test At 0.10 level of significance, critical value = -2.500

H₀: The mean time for new method is greater than or equal to 42.3 minutes H₁: The mean time for new method is less than 42.3

t = xbar - µ / s/ √ n = 40.6-42.3 / 2.7 / √24 = -3.085

P-value = 1-P(t>-3.085, df =23) = 1-0.9974 =0.0026

Since -3.085 falls in the rejection region, H₀ is rejected; or since p-value (0.0026) is less than the significance level (0.10), H₀ is rejected. At the 0.10 significance level, we can conclude that the assembly time using the new method is faster than the old method.

Lind Chapter 11; Exercise 24

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