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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 3389.194410.46611.73811.885\begin{array}{llcccc}\text { Variable } & N & \text { Lo 95\% CI } & \text { Mean } & \text { Up 95\% CI } & \text { SD } \\\text { Parking Time } & 338 & 9.1944 & 10.466 & 11.738 & 11.885\end{array} 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?


A) The university could increase their confidence level.
B) The university could decrease their confidence level.

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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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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. \[\hat { p } = \frac { 380 } { 600 } = .6333 \text { and } \hat { q } = 1 - \hat { p } = 1 - .6333 = .3667\] The confidence interval for \(p\) is \(\hat { p } \pm z \alpha / 2 \sqrt { \frac { \hat { p } \hat { q }} { n } }\). For confidence coefficient .99, \(1 - \alpha = .99 \Rightarrow \alpha = 1 - .99 = .01\). \(\alpha / 2 = .01 / 2 = .005\). \(z \alpha / 2 = z .005 = 2.575\). The \(99 \%\) confidence interval is: \[.6333 \pm 2.575 \sqrt { \frac { .6333 ( .3667 ) } { 600 } } = .6333 \pm .0507\]

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

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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.


A) We are 99% confident that the true proportion of all students receiving financial aid is between .50 and .68.
B) 99% of the students receive between 50% and 68% of their tuition in financial aid.
C) We are 99% confident that between 50% and 68% of the sampled students receive some sort of financial aid.
D) We are 99% confident that 59% of the students are on some sort of financial aid.

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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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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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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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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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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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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.


A) We are 90% confident that the mean salary of all CEOs in the electronics industry falls in the interval $139,048 to $154,144.
B) 90% of all CEOs in the electronics industry have salaries that fall between $139,048 to $154,144.
C) We are 90% confident that the mean salary of the sampled CEOs falls in the interval $139,048 to $154,144.
D) 90% of the sampled CEOs have salaries that fell in the interval $139,048 to $154,144.

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


A) .95
B) .05
C) .025
D) .475

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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 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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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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Which statement best describes a parameter?


A) A parameter is a numerical measure of a population that is almost always unknown and must be estimated.
B) A parameter is a level of confidence associated with an interval about a sample mean or proportion.
C) A parameter is a sample size that guarantees the error in estimation is within acceptable limits.
D) A parameter is an unbiased estimate of a statistic found by experimentation or polling.

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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; \( n \hat {p }\) = 8 < 15

The following data represent the scores of a sample of 50 randomly chosen students on a standardized test. 3948556366686869707171717374767676777879797979808082838383858586868888888889898990919292939596979799\begin{array} { l l l l l l l l l l } 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\end{array} 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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a. The sample mean is 79.98 an...

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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

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?


A) In construction of the confidence interval, a t-value with 19 degrees of freedom was used.
B) In construction of the confidence interval, a t-value with 20 degrees of freedom was used.
C) In construction of the confidence interval, a z-value was used.
D) In construction of the confidence interval, a z-value with 20 degrees of freedom was used. 2 Use t-Distribution

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