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

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Question 61

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The following random sample was selected from a normal population: 9, 11, 8, 10, 14, 8. Construct a 95% confidence interval for the population mean μ.

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Question 62

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Suppose you selected a random sample of n = 29 measurements from a normal distribution. Compare the standard normal z value with the corresponding t value for a 95% confidence interval.

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z: 1.96 and t: 2.048...

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Question 63

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A random sample of 15 crates have a mean weight of 165.2 pounds and a standard deviation of 14.9 pounds. Construct a 95% confidence interval for the population standard deviation σ. Assume the population is normally distributed, and round to the nearest hundredth when necessary.

Determine the confidence level for the given confidence interval for ?. $\bar { x } \pm 1.48 \left( \frac { \sigma } { \sqrt { n } } \right)$

In the construction of confidence intervals, if all other quantities are unchanged, an increase in the sample size will lead to a __________ interval.

A computer package was used to generate the following printout for estimating the mean sale price of homes in a particular neighborhood. $X=\text { sale price }$ $\begin{array}{r}\text {SAMPLE MEAN OF X =\quad 46,600 }\\\text {SAMPLE STANDARD DEV \( =13,747 \) }\\\text {SAMPLE SIZE OF X \( =15 \) }\\\text {CONFIDENCE \( =\quad 95 \) }\\\\\text {UPPER LIMIT \( =\quad 54,213.60 \) }\\\text {SAMPLE MEAN OF \( X=46,600 \) }\\\text {LOWER LIMIT \( =38,986.40 \) } \end{array}$ A friend suggests that the mean sale price of homes in this neighborhood is $44,000. Comment on your friend's suggestion.

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 90% confidence interval for the mean was calculated to be ($2,181,260, $5,836,180) . Explain what the phrase "90% confident" means.

For n = 40 and $\hat { p }$ = .35, is the sample size large enough to construct a confidence for p?

For the given combination of $\alpha$ and degrees of freedom (df), find the value of $\chi { } _ { \alpha / 2 } ^ { 2 }$ that would be used to find the lower endpoint of a confidence interval for $\sigma ^ { 2 }$ . $\alpha = 0.01 , \mathrm { df } = 30$ A) $53.6720$ B) $50.8922$ C) $52.3356$ D) $13.7867$

Find the value of $t _ { 0 } \text { such that the following statement is true: } P \left( - t _ { 0 } \leq t \leq t _ { 0 } \right) = .90 \text { where } \mathrm { df } = 14 \text {. }$

Given the values of $\bar { x } , \mathrm {~s}$ , and $\mathrm { n }$ , form a $99 \%$ confidence interval for $\sigma$ . $\overline { \mathrm { x } } = 11.8 , \mathrm {~s} = 9.8 , \mathrm { n } = 8$ A) $( 5.76,26.07 )$ B) $( 6.03,23.29 )$ C) $( 33.15,679.58 )$ D) $( 3.38,69.34 )$

How much money does the average professional football fan spend on food at a single football game? That question was posed to ten randomly selected football fans. The sampled results show that the sample mean and sample standard deviation were $70.00 and $17.50, respectively. Use this information to create a 95 percent confidence interval for the population mean. A) $70 \pm 2.228 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$ 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) $70 \pm 2.262 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$

A computer package was used to generate the following printout for estimating the mean sale price of homes in a particular neighborhood. $X=\text { sale price }$ $\begin{array}{r}\text {SAMPLE MEAN OF X =\quad 46,400 }\\\text {SAMPLE STANDARD DEV \( =13,747 \) }\\\text {SAMPLE SIZE OF X \( =15 \) }\\\text {CONFIDENCE \( =\quad 98 \) }\\\\\text {UPPER LIMIT \( =\quad 55,713.8 \) }\\\text {SAMPLE MEAN OF \( X=46,400 \) }\\\text {LOWER LIMIT \( =37.086 .2 \) } \end{array}$ At what level of reliability is the confidence interval made?

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) . Explain what the phrase "95 percent confident" means in this situation.

Given the values of $\bar { x } , s$ , and $n$ , form a $99 \%$ confidence interval for $\sigma ^ { 2 }$ . $\overline { \mathrm { x } } = 10.2 , \mathrm {~s} = 8.6 , \mathrm { n } = 8$ A) $( 25.53,523.34 )$ B) $( 28.02,417.84 )$ C) $( 29.18,598.1 )$ D) $( 3.26,48.59 )$

For quantitative data, the target parameter is most likely to be the mode of the data.

Find $\mathrm { z } _ { \alpha / 2 }$ for the given value of $\alpha$ . $\alpha = 0.14$

Explain what the phrase 95% confident means when we interpret a 95% confidence interval for μ.

A computer package was used to generate the following printout for estimating the mean sale price of homes in a particular neighborhood. $X=\text { sale price }$ $\begin{array}{r}\text {SAMPLE MEAN OF X =\quad 46,400 }\\\text {SAMPLE STANDARD DEV \( =13,747 \) }\\\text {SAMPLE SIZE OF X \( =15 \) }\\\text {CONFIDENCE \( =\quad 98 \) }\\\\\text {UPPER LIMIT \( =\quad 52,850.6 \) }\\\text {SAMPLE MEAN OF \( X=46,600 \) }\\\text {LOWER LIMIT \( =40,349.4 \) } \end{array}$ Which of the following is a practical interpretation of the interval above?

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 99% confidence interval: (.438, .642) . State the level of reliability used to create the confidence interval.