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• From Mathematics: Statistics
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• Due on Aug. 17, 2012
• Asked on Aug 14, 2012 at 1:26:18PM

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Q:

1)      Find the margin of error for the given values of c,s, and n.

c = 0.90, s = 3.2, n = 81

E =

(Round to three decimal places as needed)

2)      Construct the confidence interval for the population mean u.

c = 0.98, x = 9.6, s = 0.8, and n = 51

A 98% confidence interval for u is =

(Round to two decimal places as needed.)

3)      Construct the confidence interval for the population mean u.

c = 0.95, x = 15.4, s = 5.0, and n = 75

A 98% confidence interval for u is =

(Round to one decimal place as needed.)

4)      Use the confidence interval to find the estimated margin of error. Then find the          sample mean.

A biologist reports a confidence interval of (3.1, 3.9) when estimating the mean height (in centimeters) of a sample of seedlings.

The estimated margin of error is =

5)      Find the minimum sample size needed to estimate u for the given values c, s, and E.

c = 0.95, s = 5.2, and E = 1

Assume that a preliminary sample has atleast 30 members.

n =

(Round to the nearest whole number)

6)      You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.

A random sample of 50 home theatre systems has a mean price of \$112.00 and a standard deviation is \$15.50.

7) You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Which interval is wider? If convenient, use technology to construct the confidence intervals.

A random sample of 50 gas grills has a mean price of \$641.80 and a standard deviation of \$55.10.

8) You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Which interval is wider? If convenient, use technology to construct the confidence intervals.

A random sample of 47 eight-ounce servings of different juice drinks has a mean of 91.8 calories and a standard deviation of 43.7 calories.

9) People were polled on how many books they read the previous year. How many subjects are needed to estimate the number of books read the previous year within one book with 95% confidence? Initial survey results indicate that o= 17.2 books.

10) A doctor wants to estimate the HDL cholesterol of all 20-29 year old females. How many subjects are needed to estimate the HDL cholesterol within 4 points with 99% confidence assuming o= 14.4? Suppose the doctor would be content with 95% confidence. How does the decrease in confidence affect the sample sixe required?

11) Construct the indicated confidence interval for the population mean u using (a) a t-distribution. (b) If you had incorrectly used a normal distribution, which interval would be wider?

12) In the following situation, assume the random variable is normally distributed and use a normal or a t-distribution to construct a 90% confidence interval for the population mean. If convenient, use technology to construct the confidence interval.

(a) In a random sample of 10 adults from a nearby county, the mean waste generated per person per day was 5.46 pounds and the standard deviation was 1.28 pounds.

(b) Repeat part (a), assuming the same statistics came from a sample size of 600. Compare the results.

13) Use the given interval to find the margin of error and the sample proportion.

(0.753,0.779)

14) In a survey of 621 males ages 18-64, 398 say they have gone to the dentist in the past year.

Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. (Round all intermediate values to the nearest thousandth as needed. Round all intermediate values to the nearest thousandth as needed.)

15) In a survey of 7000 women, 4431 say they change their nail polish once a week. Construct a 99% confidence interval for the population proportion of women who change their nail polish once a week. (Round all intermediate values to the nearest thousandth as needed. Round all intermediate values to the nearest thousandth as needed.)

16) A researcher wishes to estimate, with 90% confidence, the proportion of adults who have high-speed internet access. Her estimate must be accurate within 5% of the true proportion.

a) Find the minimum sample size needed, using a prior study that found that 28% of the respondents said they have high-speed internet access.

b) No preliminary estimate is available. Find the minimum sample size needed.

( Round up to the nearest whole number as needed.)

17) The table to the right shows the results of a survey in which 2586 adults from Country A, 1138 adults from Country B, and 1066 adults from Country C were asked if human activity contributes to global warming. Complete parts (a), (b), and (c).

Adults who say that human activity contributes to global warming

Country A 64%, Country B 87%, Country C 93%

(a)    Construct a 90% confidence interval for the proportion of adults from Country A who say that human activity contributes to global warming.

(b)   Construct a 90% confidence interval for the proportion of adults from Country B who say that human activity contributes to global warming.

(c)    Construct a 90% confidence interval for the proportion of adults from Country C who say that human activity contributes to global warming.

18) The table shows the results of a survey in which separate samples of 400 adults each from the East, South,Midwest, and West were asked if traffic congestion is a serious problem in their community.

East 36%, South 33%, Midwest 26%, West 53%

(a)    Construct a 95% confidence interval for the proportion of adults from theMidwestwho say traffic congestion is a serious problem.

(b)   Construct a 95% confidence interval for the proportion of adults from the East who say traffic congestion is a serious problem.

(c)    Construct a 95% confidence interval for the proportion of adults from the South who say traffic congestion is a serious problem.

(d)   Construct a 95% confidence interval for the proportion of adults from the West who say traffic congestion is a serious problem.

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A:
Preview: ... SD/sqrt(n) = 0.449/sqrt(n). At 90% CL, z = 1.645 . E = 5% = 0.05 = z*SE = 1.645*0.449/sqrt(n) . => n  = (1.645*0.449/0.05)^2 = 218.29  = say 219 (ANSWER).   b. Let  p = 0.50 . SD = sqrt(0.50*0.50) = 0.50 . SE = SD/sqrt(n) = 0.50/sqrt(n). At 90% CL, z = 1.645 . E = 5% = 0.05 = z*SE = 1.645*0.50/sqrt(n) . => n  = (1.645*0.50/0.05)^2 = 270.60  = say 271 (ANSWER).   17.    At 90% CL, z = 1.645.   a. Country A :   SD = sqrt(p(1-p)) = sqrt(0.64*0.36) = 0.48 . SE of proportion = SD/sqrt(n) = 0.48/sqrt(2586) = 0.009439 .   90% CI = p +/- 1.645*SE , = 0.64 +/- 1.645*0.009439 , = 0.6245 to 0.6555  = ...

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