Confidence Intervals and Sample Size
Suppose the coach of the football team wants to estimate the proportion of the population of fans who support his current starter lineup. The coach wants the estimate to be .04 of the true proportion. Assume a 95 percent level of confidence. The coach estimated the proportion supporting the current starter lineup to be .60.
Answer the following questions:
1. Construct a 95% confidence interval using a sample size of 50, then of 100, then of 1,000.
2. How did changing the sample size affect the size of the interval?
3. What is the error of the estimate for each of these sample sizes?
4. How large of a sample is required for the error of the estimate to be within +.04 of the population proportion?
5. How large would the sample have to be if the coach of the team did not give an estimated proportion?
1) You manage a mail order company that advertises that they ship 85% of the orders within two working days. You take a random sample of 20 orders and check to see if indeed the order was shipped within the two day period.
a) What is the probability that 15 or more shipped within the period?
b) What is the probability that all 20 shipped within the period?
c) What is the mean of this distribution?
d) What is the standard deviation of this distribution?
e) What is the probability that less than 20 but more than 14 would be shipped?
f) What is the probability that less than 14 shipped within the period?
g) If less than 14 were shipped within the period what would your conclusion be?
h) What is the probability that between 14 and 18 of the orders were shipped within the two day period?
i) What type of probability distribution is this?
j) Explain how this fits the requirements for this type of distribution.
2) A distribution is a continuous uniform distribution between the values of 2 and 5. Draw the distribution then determine the following:
a) Mean of the distribution
b) P(2< x < 3) = ?
c) P( x > 2.5) = ?
d) P(x > 5) = ?
e) P(2< x < 5) = ?
3) In the standard normal distribution determine the following:
a) = ?
b) = ?
c) What is the probability that a value is more than 2.35 standard deviations above the mean?
d) What is the probability that a value is between -1.75 and 1.75 standard deviations from the mean?
e) What is the probability that a value is at least 2.8 standard deviations above the mean?
f) What percent of the data is below z = - 1.43?
g) What z-score represents the 38th percentile?
h) What z-score represents the 99th percentile?
i) Between what two z values will the middle 50% of the data be?
4) According to the College Board, there were 3538 students in Ohio who took at least one of the subject area SAT II tests. These students scored a mean of 651 with a standard deviation of 87 on the SAT I test. This is a normally distributed test.
a) What percent of the students scored below 500 on the test?
b) What percent of the students scored above 700 on the test?
c) Between what two values did the middle 50% of the students score?
d) What value represents the 90th percentile on this test?
e) What is the z value for a student who scored 750 on the test?
5) What are the elements that make up the sampling distribution of the sample mean?
a) How do you determine the mean of the sampling distribution of the sample mean?
b) How do you determine the standard deviation of the sampling distribution of the sample mean?
c) What is another name for this standard deviation of the sampling distribution?
d) Under what two conditions is the sampling distribution of the sample mean considered normal?
6) A population of cereal boxes has a normal weight with a mean of 16 ounces and a standard deviation of 0.2 ounces. If a box is chosen at random what is the probability that it will weigh less than 15.3 ounces? If a sample of 9 boxes is randomly chosen, what is the probability that the mean will be less than 15.3 ounces? If a sample of 25 boxes is randomly chosen, what is the probability that the mean will be less than 15.3 ounces? If you were in charge of filling the boxes and the sample of 25 boxes that you chose had a mean of 15.3 ounces what action would you take?© BrainMass Inc. brainmass.com October 25, 2018, 8:34 am ad1c9bdddf
The solution provides step by step method for the calculation of confidence interval and sample size. Formula for the calculation and Interpretations of the results are also included.
Statistics: Confidence intervals or large and small samples, population proportions
Section 6.1: Confidence Intervals for the Mean (Large Samples)
1. Find the critical value zc necessary to form a confidence interval at the given level of confidence. (References: definition for level of confidence page 311, end of section exercises 5 - 8 page 317)
2. In a random sample of 60 computers, the mean cost for repairs was $150. From past studies, it was found that the standard deviation was  = $36.
(References: example 5 page 315, end of section exercises 51 - 56 pages 319 - 320)
a. Find the margin of error E for a 90% confidence interval.
Round your answer to the nearest hundredths. . (References:
definition of margin of error on page 312 and example 2 on
b. Construct a 90% confidence interval for the mean life,  of repair costs.
3. A nurse at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she desires to be 98% confident that the true mean is within 3 ounces of the sample mean? The standard deviation  is known to be 6 ounces. (References: example 6 page 316, end of section exercises 58 - 62 pages 321 - 322)
Section 6.2: Confidence Intervals for the Mean (Small Samples)
4. A random sample of 16 fluorescent light bulbs has a mean life of 645 hours with a sample standard deviation of 31 hours. Assume the population has a normal distribution.(References: example 2 and 3 pages 327 - 328, end of section exercises 5 - 16 pages 330 - 331)
a. Find the margin of error for a 95% confidence interval. Round your answer to the nearest tenths.
b. Find a 95% confidence interval for the mean  for all fluorescent light bulbs.
Section 6.3: Confidence Intervals for Population Proportions
5. In a survey of 2480 golfers, 15% said they were left-handed. Construct to the 95% confidence interval for the population proportion p. (References: example 1 - 3 page 334 - 337, end of section exercises 13 - 20 page 339 - 340)
a. Find the margin of error E.
Round E to three decimal places
c. Construct a 95% confidence interval for the population proportion p of left-handed golfers.View Full Posting Details