Problem Set 2
a. Lind Chapter 9: Exercises 12, 28
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 the population mean is unknown. But we can use the sample mean to estimate it. So, the best estimate of this value is 60 pounds.
b. Explain why we need to use the t distribution. What assumption do you need to make?
Since the standard deviation of the population is unknown, we need to use the distribution. We need to assume that the population follows normal distribution.
c. For a 90 percent confidence interval, what is the value of t?
For a 90 percent confidence interval, the value of t is 2.131, we use the degree of freedom df=n-1=16-1=15.
d. Develop the 90 percent confidence interval for the population mean.
i.e.,
(49.345, 70.655)
e. Would it be reasonable to conclude that the population mean is 63 pounds?
Yes, since 63 is in the above 90% confidence interval.
28. Suppose the President wants an estimate of the proportion of the population who support his current policy toward gun control. The President wants the estimate to be within .04 of the true proportion. Assume a 95 percent level of confidence. The President's political advisors estimated the proportion supporting the current policy to be .60.
a. How large of a sample is required?
Formula for sample size
n = (z^2pq)/d^2
Given z = 1.96 (95% CI)
p = 0.60
q = 1 - p = 0.40
d = 0.04
n = [(1.96^2)(0.6)(0.4)]/...
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