1. ## Population Problems

1.) A recent study of 100 individuals revealed that in the US the mean amount of money spent on gifts for Mother’s Day for a given individual is $38. The population standard deviation is not known but is estimated to be$12.
a.) Using a 90% confidence, what is the margin of error in estimating the mean?
b.) Using your results from part (a), what is the 90% confidence interval for the mean? Interpret your results.
c.) How many individuals need to be sampled in order to have a margin of error of $1.2 at 90% confidence 2.) Suppose a survey reveals that 12 out of 1500 individuals are a twin. a.) What is the point estimate of the population proportion? b.) Using a 95% confidence, what is the margin of error in estimating the population proportion? c.) Using your results from part (b), what is the 95% confidence interval for the population proportion? 3.) In the United States, the mean monthly electric bill is$250 per household. A sample of 100 households is taken and showed a sample mean of $247.60. Suppose the population standard deviation is$10.00. Suppose you wish to determine if the mean monthly electric bill is under $250 per household. Assume the probability of a Type I error is 0.05. a.) State the null and alternative hypotheses. b.) What is the value of the test statistic? c.) What is the p-value? d.) State your conclusions. 2. Here's what I've got, is it right? 1.) A recent study of 100 individuals revealed that in the US the mean amount of money spent on gifts for Mother’s Day for a given individual is$38. The population standard deviation is not known but is estimated to be $12. a.) Using a 90% confidence, what is the margin of error in estimating the mean? • α = 1 – 0.90 = 0.10 • p* = 1 – (0.10/2) = 0.95 ---> critical value of: 1.65 • Margin of Error: 1.65 * 12/sqrt(100) = 1.98 b.) Using your results from part (a), what is the 90% confidence interval for the mean? Interpret your results. • 38 +/- 1.98 or 36.02 to 39.98 or 36 to 40 c.) How many individuals need to be sampled in order to have a margin of error of$1.2 at 90% confidence?

2.) Suppose a survey reveals that 12 out of 1500 individuals are a twin.
a.) What is the point estimate of the population proportion?
• 12 / 15000 = .008 or .8%

b.) Using a 95% confidence, what is the margin of error in estimating the population proportion?
• α = 1 – 0.95 = 0.05
• 1.96 x sqrt([(0.008)(0.992)]/1500) = 1.96 x 0.0023 = 0.004508

c.) Using your results from part (b), what is the 95% confidence interval for the population proportion?
• .008 +/- 0.004508 or .003492 to .012508

3.) In the United States, the mean monthly electric bill is $250 per household. A sample of 100 households is taken and showed a sample mean of$247.60. Suppose the population standard deviation is $10.00. Suppose you wish to determine if the mean monthly electric bill is under$250 per household. Assume the probability of a Type I error is 0.05.
a.) State the null and alternative hypotheses.
• H0: μ < 250
• H1: μ > 250
b.) What is the value of the test statistic?
• z = [250 – 247.60] / [10/sqrt(100)] = 2.4
c.) What is the p-value?
• For z = 2.4, Cumulative prob = .9932
• 1-.9932 = .0068