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Hi, here're the questions (looks like there are 8 questions, not 9):

#1: Survey asked, “During the last year, did anyone take something from you by using force – such as a stickup, mugging, or threat? Of 962 subjects, 20 answered yes and 929 no.

I have already found the point estimate of the proportion of the populations who were victims – it is 0.02079
Also need to be found:
a. standard error of the estimate (round to 5 decimal places)
b. margin of error for 95% confidence interval (round to 5 decimal places)
c. construct confidence interval for the population proportion (use ascending order, round to 5 decimal places)

#2: Results of survey on belief in reincarnation are shown in the table below:

X= 592
N=2212
Sample p=0.267631
95% CI=(0.249181, 0.286081)

a. using confidence interval shown, find the 95% confidence interval for the population proportion who don’t believe in reincarnation

#3: Survey asks, “If husband in a family wants kids, but the wife decides she doesn’t want any, is it all rights for the wife to refuse to have kids? Of 719 subjects, 591 said yes.

a. Find 99% confidence interval for the population proportion who would say yes (ascending order, round to 5 decimal places)
b. Can you conclude that population proportion exceeds 75%? Why?

#4: Survey asked the question, “What do you think is the ideal # of kids for a family to have?” 504 females who responded has a median of 2, mean of 3.11 and standard deviation of 1.26.
a. What is point estimate of population mean?
b. Find standard error of population mean.

#5: Selling price of a PDA in 18 randomly chosen stores are:
251 245 257 206 214 242 229 257 233 245 216 176 251 242 238 229 245 229
These data have x=233.61, s=20.42, Q1= 229, Median=240, Q3=245.

a. determine 95% confidence interval for the population mean (lower limit, upper limit – round to 2 decimal places).
b. The dotplot shows that $176 observation is quite a bit lower than the others. Deleting this observation, find the mean & standard deviation, & construct the 95% confidence interval for mean.

#6: When survey asked, “About how many hours per week do you spend sending and answering email?”, 7 females in the survey sample of age at least 80 had the responses 0, 0, 3, 4, 5, 7, 14.

a. using software or calculator, find the sample mean & standard deviation & standard error of sample mean (round to the nearest hundredth)
b. find a 90% confidence interval for the population mean (round to the nearest tenth)


#7: Survey asked 6502 random adults the question “How any children have you ever had?” Use the result below to answer the questions

# of kids 0 1 2 3 4 5 6 7 8
N 1985 969 1437 890 453 334 230 140 64

a. find standard error of the sample mean f the random sample of 6502 adults.
b. Construct a 95% confidence interval for the population mean.

#8: Students at a certain school were surveyed, and it was estimated that 29% of college students abstain from drinking alcohol. To estimate this proportion in your school, how large a random sample would you need to estimate it to within 0.05 with probability 0.99, if before conducting the study
(a) you are unwilling to predict the proportion value at your school (n=?; round to nearest integer)
(b) you use the results from the surveyed school as a guideline (n=?; round to nearest integer).

Thanks!