Posted by **penelope** on Friday, July 12, 2013 at 12:54am.

I am having a lot of trouble with these questions can anyone help me please.

2. On September 2, 2004, a Dan Jones poll based on a random sample of 408 Salt Lake County residents reported that 26% of the respondents thought that Mayor Nancy Workman should resign. Find a 95% confidence interval for the percent of all residents in Salt Lake County that felt Nancy Workman should resign.

a. 22% to 30%

b. 24% to 28%

c. 70% to 78%

d. 17% to 35%

3. A health center noted that in a simple random sample of 600 patients, 120 were referred to them by the local hospital. An approximate 90% confidence interval for the percentage of health center patients who were referred by the hospital is:

a. 17.31% to 22.69%

b. 13.4% to 26.6%

c. 16.73% to 23.27%

d. 15.10% to 24.90%

4. A candy company is market-testing a new brand of chocolate bar. A simple random sample of 400 people were given the new chocolate bar to try. A total of 285 said they liked the candy. Choose the statement that is FALSE.

a. The sample percentage is 71.25%; the SE for the sample percentage is 2.26%.

b. 66.73% to 75.77% is a 95% confidence interval for the population percentage of people who will like the new chocolate bar.

c. 66.73% to 75.77% is a 95% confidence interval for the sample percentage of people who will like the new chocolate bar.

d. The candy company can be 95% confident that the true percentage of consumers who will like their new chocolate bar is between 66.73% and 75.77%.

5. A box of tickets has an average of 32 with an SD of 6. Two hundred draws are made at random with replacement from this box. Find the approximate chance that the average of the draws will be between 31.3 and 32.7.

a. 8%

b. 16%

c. 90%

d. 50%

- college stats -
**PsyDAG**, Friday, July 12, 2013 at 11:28am
2. 95% = mean ± 1.96 SEm

SEm = SD/√n

3. 90% = mean ± 1.645 SEm

5. Z = (score-mean)/SEm

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z scores. The values for the SEm (1.96, !.645) on previous problems were also obtained from this table.

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