Exam 7: Inferences Based on a Single Sample: Estimation With Confidence Intervals

Since the population standard deviation σ is almost always known, we use it instead of the sample standard deviation s when finding a confidence interval.

How much money does the average professional football fan spend on food at a single football game? That question was posed to 40 randomly selected football fans. The sample results provided a sample mean and standard deviation of $11.00 and $2.80, respectively. Find and interpret a 99% confidence interval for μ.

A previous random sample of 4000 U.S. citizens yielded 2250 who are in favor of gun control legislation. How many citizens would need to be sampled for a 99% confidence interval to estimate the true proportion within 5%?

An educator wanted to look at the study habits of university students. As part of the research, data was collected for three variables - the amount of time (in hours per week) spent studying, the amount of time (in hours per week) spent playing video games and the GPA - for a sample of 20 male university students. As part of the research, a 95% confidence interval for the average GPA of all male university students was calculated to be: (2.95, 3.10). What assumption is necessary for the confidence interval analysis to work properly?

To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has recently been marketed. The FDA tests on this cigarette yielded a mean nicotine content of 24.5 milligrams and standard deviation of 2.3 milligrams for a sample of n = 82 cigarettes. Find a 95% confidence interval for μ.

The daily intakes of milk (in ounces) for ten five-year old children selected at random from one school were: 28.3 16.6 23.8 25.7 23.0 13.2 23.4 22.1 13.8 10.7 Find a 99% confidence interval for the standard deviation, ?, of the daily milk intakes of all five-year olds at this school. Round to the nearest hundredth when necessary.

Suppose (1,000, 2,100) is a 95% confidence interval for μ. To make more useful inferences from the data, it is desired to reduce the width of the confidence interval. Explain why an increase in sample size will lead to a narrower interval of the estimate of μ.

A random sample of n measurements was selected from a population with unknown mean μ and known standard deviation σ. Calculate a 95% confidence interval for μ for the given situation. Round to the nearest hundredth when necessary. $n = 100 , \bar { x } = 58 , \sigma = 20$ A) $58 \pm 3.92$ B) $58 \pm 3.29$ C) $58 \pm 39.2$ D) $58 \pm 0.39$

A confidence interval was used to estimate the proportion of statistics students who are female. A random sample of 72 statistics students generated the following confidence interval: (.438, .642). Using the information above, what sample size would be necessary if we wanted to estimate the true proportion to within 1% using 95% reliability?

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. If the dean wanted to estimate the proportion of all students receiving financial aid to within 3% with 90% reliability, how many students would need to be sampled?

A marketing research company is estimating which of two soft drinks college students prefer. A random sample of 157 college students produced the following confidence interval for the proportion of college students who prefer drink A: (.344, .494). Is this a large enough sample for this analysis to work? A) Yes, since both $\hat { n p } \geq 15$ and $\hat { n q } \geq 15$ . B) Yes, since $n = 157$ (which is 30 or more). C) No. D) It is impossible to say with the given information.

Suppose that 100 samples of size n = 50 are independently chosen from the same population and that each sample is used to construct its own 95% confidence interval for an unknown population mean μ. How many of the 100 confidence intervals would you expect to actually contain μ?

How much money does the average professional football fan spend on food at a single football game? That question was posed to 60 randomly selected football fans. The sampled results show that the sample mean was $70.00 and prior sampling indicated that the population standard deviation was $17.50. Use this information to create a 95 percent confidence interval for the population mean. A) $70 \pm 1.645 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$ B) $70 \pm 1.960 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$ C) $70 \pm 1.833 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$ D) $70 \pm 1.671 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$

You are interested in purchasing a new car. One of the many points you wish to consider is the resale value of the car after 5 years. Since you are particularly interested in a certain foreign sedan, you decide to estimate the resale value of this car with a 99% confidence interval. You manage to obtain data on 17 recently resold 5-year-old foreign sedans of the same model. These 17 cars were resold at an average price of $12,380 with a standard deviation of $800. What is the 99% confidence interval for the true mean resale value of a 5- year-old car of this model? A) $12,380 \pm 2.921 ( 800 / \sqrt { 17 } )$ B) $12,380 \pm 2.898 ( 800 / \sqrt { 17 } )$ C) $12,380 \pm 2.575 ( 800 / \sqrt { 17 } )$ D) $12,380 \pm 2.921 ( 800 / \sqrt { 16 } )$

A retired statistician was interested in determining the average cost of a $200,000.00 term life insurance policy for a 60-year-old male non-smoker. He randomly sampled 65 subjects (60-year-old male non-smokers) and constructed the following 95 percent confidence interval for the mean cost of the term life insurance: ($850.00, $1050.00). What value of alpha was used to create this confidence interval?

A marketing research company is estimating the average total compensation of CEOs in the service industry. Data were randomly collected from 18 CEOs and the 97% confidence interval was calculated to be ($2,181,260, $5,836,180). Give a practical interpretation of the confidence interval.

Find the value of $t _ { 0 } \text { such that the following statement is true: } P \left( - t _ { 0 } \leq t \leq t _ { 0 } \right) = .99 \text { where } \mathrm { df } = 9 \text {. }$

A computer package was used to generate the following printout for estimating the mean sale price of homes in a particular neighborhood. $X=\text { sale price }$ SAMPLE MEAN OF X =46300 SAMPLE STANDARD DEV =13747 SAMPLE SIZE OF X =25 CONFIDENCE =90 UPPER LIMIT =51003.90 SAMPLE MEAN OF X=46300 LOWER LIMIT =31596.10 A friend suggests that the mean sale price of homes in this neighborhood is $42,000. Comment on your friend's suggestion.

A newspaper reports on the topics that teenagers most want to discuss with their parents. The findings, the results of a poll, showed that 46% would like more discussion about the family's financial situation, 37% would like to talk about school, and 30% would like to talk about religion. These and other percentages were based on a national sampling of 505 teenagers. Estimate the proportion of all teenagers who want more family discussions about religion. Use a 99% confidence level.

The following sample of 16 measurements was selected from a population that is approximately normally distributed. 61 85 92 77 83 81 75 78 95 87 69 74 76 84 80 83 Construct a 90% confidence interval for the population mean.