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

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The director of a hospital wishes to estimate the mean number of people who are admitted to the emergency room during a 24-hour period. The director randomly selects 36 different 24-hour periods and determines the number of admissions for each. For this sample, xˉ=17.3\bar { x } = 17.3 and s2=16s ^ { 2 } = 16 . Estimate the mean number of admissions per 24 -hour period with a 99%99 \% confidence interval. A) 17.3±1.71717.3 \pm 1.717 B) 17.3±6.86717.3 \pm 6.867 C) 17.3±.28617.3 \pm .286 D) 17.3±.66017.3 \pm .660

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A

The U.S. Commission on Crime randomly selects 600 files of recently committed crimes in an area and finds 380 in which a firearm was reportedly used. Find a 99% confidence interval for p, the true fraction of crimes in the area in which some type of firearm was reportedly used.

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Let p = the true fraction of crimes in the area in which some type of firearm was reportedly used. p^=380600=.6333 and q^=1p^=1.6333=.3667\hat { p } = \frac { 380 } { 600 } = .6333 \text { and } \hat { q } = 1 - \hat { p } = 1 - .6333 = .3667
The confidence interval for pp is p^±zα/2p^q^n\hat { p } \pm z \alpha / 2 \sqrt { \frac { \hat { p } \hat { q }} { n } } .
For confidence coefficient .99, 1α=.99α=1.99=.011 - \alpha = .99 \Rightarrow \alpha = 1 - .99 = .01 . α/2=.01/2=.005\alpha / 2 = .01 / 2 = .005 .

zα/2=z.005=2.575z \alpha / 2 = z .005 = 2.575 . The 99%99 \% confidence interval is:
.6333±2.575.6333(.3667)600=.6333±.0507.6333 \pm 2.575 \sqrt { \frac { .6333 ( .3667 ) } { 600 } } = .6333 \pm .0507

For n = 800 and p^\hat { p } = .99, is the sample size large enough to construct a confidence for p?

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No; np^ n \hat {p }

= 8 < 15

Parking at a large university can be extremely difficult at times. One particular university is trying to determine the location of a new parking garage. As part of their research, officials are interested in estimating the average parking time of students from within the various colleges on campus. A survey of 338 College of Business (COBA) students yields the following descriptive information regarding the length of time (in minutes) it took them to find a parking spot. Note that the "Lo 95%" and "Up 95%" refer to the endpoints of the desired confidence interval. Variable N Lo 95\% CI Mean Up 95\% CI SD Parking Time 338 9.1944 10.466 11.738 11.885 University officials have determined that the confidence interval would be more useful if the interval were narrower. Which of the following changes in the confidence level would result in a narrower interval?

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The confidence level is the confidence coefficient expressed as a percentage.

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A random sample of 80 observations produced a mean xˉ=35.4 and a standard deviation s=3.1\bar { x } = 35.4 \text { and a standard deviation } s = 3.1 \text {. } a. Find a 90% confidence interval for the population mean μ. b. Find a 95% confidence interval for μ. c. Find a 99% confidence interval for μ. d. What happens to the width of a confidence interval as the value of the confidence coefficient is increased while the sample size is held fixed? 7.3 Confidence Interval for a Population Mean: Student's t-Statistic 1 Compare t-Distribution to Normal Distribution

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A random sample of 4000 U.S. citizens yielded 2250 who are in favor of gun control legislation. Estimate the true proportion of all Americans who are in favor of gun control legislation using a 95% confidence interval. A) .5625±.0154.5625 \pm .0154 B) .5625±.4823.5625 \pm .4823 C) .4375±.0154.4375 \pm .0154 D) .4375±.4823.4375 \pm .4823

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The following data represent the scores of a sample of 50 randomly chosen students on a standardized test. 39 48 55 63 66 68 68 69 70 71 71 71 73 74 76 76 76 77 78 79 79 79 79 80 80 82 83 83 83 85 85 86 86 88 88 88 88 89 89 89 90 91 92 92 93 95 96 97 97 99 a. Write a 95% confidence interval for the mean score of all students who took the test. b. Identify the target parameter and the point estimator.

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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). Which of the following statements is true?

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Which statement best describes a parameter?

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One way of reducing the width of a confidence interval is to reduce the size of the sample taken.

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A newspaper reported 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 549 teenagers. Using 99% reliability, can we say that more than 30% of all teenagers want to discuss school with their parents? A) Yes, since the values inside the 99% confidence interval are greater than .30. B) No, since the value .30 is not contained in the 99% confidence interval. C) Yes, since the value .30 falls inside the 99% confidence interval. D) No, since the value .30 is not contained in the 99% confidence interval.

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What is the confidence coefficient in a 95% confidence interval for μ?

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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. The 99% confidence interval for p is 59 ± .09. Interpret this interval.

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A random sample of n = 144 measurements was selected from a population with unknown mean μ and standard deviation σ. Calculate a 90% confidence interval if xˉ=3.55 and s=.49\bar { x } = 3.55 \text { and } s = .49

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What type of car is more popular among college students, American or foreign? One hundred fifty-nine college students were randomly sampled and each was asked which type of car he or she prefers. A computer package was used to generate the printout below for the proportion of college students who prefer American automobiles. SAMPLE PROPORTION = .396226 SAMPLE SIZE = 159 UPPER LIMIT = .46418 LOWER LIMIT = .331125 Is the sample large enough for the interval to be valid? A) Yes, since np^n \hat{p} and nq^n \hat{q} are both greater than 15 . B) Yes, since n>30n > 30 . C) No, the sample size should be at 10%10 \% of the population. D) No, the population of college students is not normally distributed.

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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 mean nicotine content of 28.4 milligrams and standard deviation of 2.2 milligrams for a sample of n = 9 cigarettes. Construct a 98% confidence interval for the mean nicotine content of this brand of cigarette. A) 28.4±2.12428.4 \pm 2.124 B) 28.4±2.06928.4 \pm 2.069 C) 28.4±2.25328.4 \pm 2.253 D) 28.4±2.19428.4 \pm 2.194

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A 90% confidence interval for the average salary of all CEOs in the electronics industry was constructed using the results of a random survey of 45 CEOs. The interval was ($139,048, $154,144). Give a practical interpretation of the interval.

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How much money does the average professional football fan spend on food at a single football game? That question was posed to 10 randomly selected football fans. The sample results provided a sample mean and standard deviation of $19.00 and $2.95, respectively. Use this information to construct a 98% confidence interval for the mean. A) 19±2.821(2.95/10)19 \pm 2.821 ( 2.95 / \sqrt { 10 } ) B) 19±2.764(2.95/10)19 \pm 2.764 ( 2.95 / \sqrt { 10 } ) C) 19±2.718(2.95/10)19 \pm 2.718 ( 2.95 / \sqrt { 10 } ) D) 19±2.262(2.95/10)19 \pm 2.262 ( 2.95 / \sqrt { 10 } )

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If no estimate of p exists when determining the sample size for a confidence interval for a proportion, we can use .5 in the formula to get a value for n.

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