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 Mothers 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?

About 270

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

d.) State your conclusions.

Since H0 = .0068 < 0.05, we reject H0