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Exam 9: Estimating the Value of a Parameter

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Question 21

Multiple Choice

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An article a Florida 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 familyʹs financialsituation, 37% would like to talk about school, and 30% would like to talk about religion. These and otherpercentages were based on a national sampling of 546 teenagers. Estimate the proportion of all teenagers who^want more family discussions about school. Use a 95% confidence level. Express the answer in the form p ± Eand round to the nearest thousandth.

A) 0.37 ± 0.040
B) 0.37 ± 0.002
C) 0.63 ± 0.040
D) 0.63 ± 0.002

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Question 22

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A survey of 100 fatal accidents showed that in 43 cases the driver at fault was inadequately insured. Find apoint estimate for p, the population proportion of accidents where the driver at fault was inadequately insured

A) 0.43
B) 0.57
C) 0.754
D) 0.301

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Question 23

Multiple Choice

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A confidence interval was used to estimate the proportion of math majors that are female. A random sample of72 math majors generated the following confidence interval: (0.438, 0.642) . Using the information above, whatsize sample would be necessary if we wanted to estimate the true proportion to within 3% using 98%reliability?

A survey of 700 non-fatal accidents showed that 143 involved faulty equipment. Find a point estimate for p,the population proportion of accidents that involved faulty equipment.

$\text { True or False: The general form of a large-sample } ( 1 - \alpha ) 100 \% \text { confidence interval for a population proportion }$ interest.

A senator wishes to estimate the proportion of United States voters who favor abolishing the Electoral College.How large a sample is needed in order to be 99% confident that the sample proportion will not differ from thetrue proportion by more than 3%?

Find the critical t-value that corresponds to 99% confidence and n = 10.

True or False: As the number of degrees of freedom increases, the chi-square distribution becomes morenearly symmetric.

To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes theamount of nicotine found in all commercial brands of cigarettes. A new cigarette has recently been marketed.The FDA tests on this cigarette gave a mean nicotine content of 28.4 milligrams and standard deviation of 2.6milligrams for a sample of n = 9 cigarettes. The FDA claims that the mean nicotine content exceeds 31.0milligrams for this brand of cigarette, and their stated reliability is 90%. Do you agree?

Assume that the heights of bookcases are normally distributed. A random sample of 16 bookcases in onecompany have a mean height of 67.5 inches and a standard deviation of 2.1 inches. Construct a 99% confidenceinterval for the population standard deviation, $\sigma _ { . }$

The table shows the monthly rents (in dollars) for 10 studio apartments selected randomly from all studioapartments in one city. $\begin{array} { r r r r r } 950 & 840 & 1330 & 1020 & 1180 \\\hline 880 & 990 & 760 & 1250 & 825\end{array}$ Explain the algorithm in using the bootstrap method with 1000 resamples to obtain a 90% confidence intervalfor the mean monthly rent of all studio apartments in the city.

True or False: The chi-square distribution is a symmetric distribution for all degrees of freedom.

What effect will an outlier have on a confidence interval that is based on a small sample size?

A computer package was used to generate the following printout for estimating the sale price of condominiumsin a particular neighborhood. $\begin{array}{rl}\text { X}=&\text { sale-price }\\\\\text { SAMPLE MEAN OF X} =&46,600\\\text { SAMPLE STANDARD DEV }=&13,747\\\text { SAMPLE SIZE OF X }= &15\\\text { CONFIDENCE } = &99\\\\\text { UPPER LIMIT }= & 57,166.70 \\\text { SAMPLE MEAN OF X }= & 46,600 \\\text { LOWER LIMIT }= & 36,033.30\end{array}$ What assumptions are necessary for any inferences derived from this printout to be valid?

Let t0 be a specific value of t. Find t0 such that the statement is true: $\mathrm { P } \left( \mathrm { t } \geq \mathrm { t } _ { 0 } \right) = 0.005 \text { where } \mathrm { df } = 20$

In a survey of 10 musicians, 2 were found to be left-handed. Is it practical to construct the 90% confidenceinterval for the population proportion, p? Explain.

Many people think that a national lobbyʹs successful fight against gun control legislation is reflecting the will ofa minority of Americans. A random sample of 4000 citizens yielded 2250 who are in favor of gun controllegislation. Estimate the true proportion of all Americans who are in favor of gun control legislation using a^99% confidence interval. Express the answer in the form p ± E and round to the nearest ten-thousandth.

88%