An inventor has developed a new spray coating that is designed to improve the wear of bicycle tires. To test the new coating, the inventor randomly selects one of the two tires on each of 50 bicycles to be coated with the new spray. The bicycle is then driven for 100 miles and the amount of the depth of the tread left on the two bicycle tires is measured (in millimeters). It is desired to determine whether the new spray coating improves the wear of the bicycle tires. The data and summary information is shown below: \[\begin{array}{l} \begin{array} { c | c | c } \text { Bicycle } & \text { Coated Tire } ( \mathrm { C } ) & \text { Non-Coated Tire (N) } \\ \hline 1 & 1.452 & 0.785 \\ 2 & 1.634 & 0.844 \\ \downarrow & \downarrow & \downarrow \\ 50 & 1.211 & 0.954 \end{array}\\\\ \begin{array} { l l c c } & \text { Coated } & \text { Non-Coated } & \text { Difference } \\ \text { Mean } & 1.38 & 0.85 & 0.53 \\ \text { Std. Dev. } & 0.12 & 0.11 & 0.06 \\ \text { Sample Size } & 50 & 50 & 50 \end{array} \end{array}\] Identify the correct null and alternative hypothesis for testing whether the new spray coating improves the mean wear of the bicycle tires (which would result in a larger amount of tread left on the tire).

A) $\mathrm { H } _ { 0 } : \mu _ { \mathrm { d } } = 0$ vs. $\mathrm { H } _ { \mathrm { a } } : \mu _ { \mathrm { d } } 0$ A) $\mathrm { H } _ { 0 } : \mu _ { \mathrm { d } } = 0$ vs. $\mathrm { H } _ { \mathrm { a } } : \mu _ { \mathrm { d } } 0$ C

In a controlled laboratory environment, a random sample of 10 adults and a random sample of 10 children were tested by a psychologist to determine the room temperature that each person finds most comfortable. The data are summarized below: \[\begin{array} { l c c } \hline & \text { Sample Mean } & \text { Sample Variance } \\ \hline \text { Adults (1) } & 77.5 ^ { \circ } \mathrm { F } & 4.5 \\ \text { Children (2) } & 74.5 ^ { \circ } \mathrm { F } & 2.5 \\ \hline \end{array}\] Suppose that the psychologist decides to construct a 99% confidence interval for the difference in mean comfortable room temperatures instead of proceeding with a test of hypothesis. The 99% confidence interval turns out to be (-2.9, 3.1). Select the correct statement.

A) It can be concluded at the 99% confidence level that the true mean comfortable room temperature for children exceeds that for adults. B) It can be concluded at the 99% confidence level that the true mean room temperature for adults exceeds that for children. C) It can be concluded at the 99% confidence level that the true mean comfortable room temperature is between -2.9 and 3.1. D) It cannot be concluded at the 99% confidence level that there is actually a difference between the true mean comfortable room temperatures for the two groups. A) It can be concluded at the 99% confidence level that the true mean comfortable room temperature for children exceeds that for adults. B) It can be concluded at the 99% confidence level that the true mean room temperature for adults exceeds that for children. C) It can be concluded at the 99% confidence level that the true mean comfortable room temperature is between -2.9 and 3.1. D) It cannot be concluded at the 99% confidence level that there is actually a difference between the true mean comfortable room temperatures for the two groups.

The FDA is comparing the mean caffeine contents of two brands of cola. Independent random samples of 6-oz. cans of each brand were selected and the caffeine content of each can determined. the study provided the following summary information. $\begin{array}{lrr} \hline & \text { Brand A } & \text { Brand B } \\ \hline \text { Sample size } & 15 & 10 \\ \text { Mean } & 18 & 20 \\ \text { Variance } & 1.2 & 1.5 \\ \hline \end{array}$ How many cans of each soda would need to be sampled in order to estimate the difference in the mean caffeine content to within . 5 with $95 \%$ reliability?

A) $n _ { 1 } = n _ { 2 } = 21$ B) $n _ { 1 } = n _ { 2 } = 57$ C) $n _ { 1 } = n _ { 2 } = 42$ D) $n _ { 1 } = n _ { 2 } = 18$ A) $n _ { 1 } = n _ { 2 } = 21$ B) $n _ { 1 } = n _ { 2 } = 57$ C) $n _ { 1 } = n _ { 2 } = 42$ D) $n _ { 1 } = n _ { 2 } = 18$

Consider the following set of salary data: $\begin{array}{lrr} \hline & \text { Men (1) } & \text { Women (2) } \\ \hline \text { Sample Size } & 100 & 80 \\ \text { Mean } & \$ 12,850 & \$ 13,000 \\ \text { Standard Deviation } & \$ 345 & \$ 500 \\ \hline \end{array}$ Suppose the test statistic turned out to be $z = - 1.20$ (not the correct value). Find a two-tailed $p$-value for this test statistic.

A) .2302 B) .1151 C) .3849 D) .6151 A) .2302 B) .1151 C) .3849 D) .6151

In a controlled laboratory environment, a random sample of 10 adults and a random sample of 10 children were tested by a psychologist to determine the room temperature that each person finds most comfortable. The data are summarized below: $\begin{array}{lcc} \hline & \text { Sample Mean } & \text { Sample Variance } \\ \hline \text { Adults (1) } & 77.5^{\circ} \mathrm{F} & 4.5 \\ \text { Children (2) } & 74.5^{\circ} \mathrm{F} & 2.5 \\ \hline \end{array}$ If the psychologist wished to test the hypothesis that children prefer warmer room temperatures than adults, which set of hypotheses would he use?

A) $H _ { 0 } : \left( \mu _ { 1 } - \mu _ { 2 } \right) = 0$ vs. $H _ { 0 } : \left( \mu _ { 1 } - \mu _ { 2 } \right) > 0$ B) $H _ { 0 } : \left( \mu _ { 1 } - \mu _ { 2 } \right) = 3$ vs. $H _ { 0 } : \left( \mu _ { 1 } - \mu _ { 2 } \right) \neq 0$ C) $H _ { 0 } : \left( \mu _ { 1 } - \mu _ { 2 } \right) = 0$ vs. $H _ { 0 } : \left( \mu _ { 1 } - \mu _ { 2 } \right) = 0$ D) $H _ { 0 } : \left( \mu _ { 1 } - \mu _ { 2 } \right) = 0$ vs. $H _ { 0 } : \left( \mu _ { 1 } - \mu _ { 2 } \right) 0$ B) $H _ { 0 } : \left( \mu _ { 1 } - \mu _ { 2 } \right) = 3$ vs. $H _ { 0 } : \left( \mu _ { 1 } - \mu _ { 2 } \right) \neq 0$ C) $H _ { 0 } : \left( \mu _ { 1 } - \mu _ { 2 } \right) = 0$ vs. $H _ { 0 } : \left( \mu _ { 1 } - \mu _ { 2 } \right) = 0$ D) $H _ { 0 } : \left( \mu _ { 1 } - \mu _ { 2 } \right) = 0$ vs. $H _ { 0 } : \left( \mu _ { 1 } - \mu _ { 2 } \right) < 0$

A cola manufacturer invited consumers to take a blind taste test. Consumers were asked to decide which of two sodas they preferred. The manufacturer was also interested in what factors played a role in taste preferences. Below is a printout comparing the taste preferences of men and women. HYPOTHESIS: PROP. $X =$ PROP. $Y$ SAMPLES SELECTED FROM soda(brand1, brand2) $\begin{array}{lll} \text { males } & (\text { sex }=0, \text { males }) & (\text { NUMBER }=115) \\ \text { females } & (\text { sex }=1, \text { females }) & (\text { NUMBER }=56) \end{array}$ $ X= $ males $ \mathrm{Y}= $ females SAMPLE PROPORTION OF X $= 0.422018$ SAMPLE SIZE OF $X = 109$ SAMPLE PROPORTION OF $Y = 0.25$ SAMPLE SIZE OF $\mathrm { Y } = 52$ PROPORTION X - PROPORTION Y = $0.172018$ \[Z = 2.11825\] Suppose the manufacturer wanted to test to determine if the males preferred its brand more than the females. Using the test statistic given, compute the appropriate p-value for the test.

A) .2119 B) .0340 C) .0170 D) .4681 A) .2119 B) .0340 C) .0170 D) .4681

Salary data were collected from CEOs in the consumer products industry and CEOs in the telecommunication industry. The data were analyzed using a software package in order to compare mean salaries of CEOs in the two industries. $\text { HYPOTHESIS: MEAN X = MEAN Y }$ $\text { SAMPLES SELECTED FROM SALARY }$ $\begin{array}{rc} &\text { \( X= $ Consumer Products}\\ &\text { $ \mathrm{Y}= $ Telecommunications}\\ \\ \text { SAMPLE MEAN OF } X= & 1761 \\ \text { SAMPLE VARIANCE OF X }= & 3.97555 \text { E } 6 \\ \text { SAMPLE SIZE OF X }= & 21 \\ \text { SAMPLE MEAN OF Y }= & 1093.5 \\ \text { SAMPLE VARIANCE OF Y }= & 103255 \\ \text { SAMPLE SIZE OF Y }= & 21\\ \\ \text { MEAN X - MEAN Y }= & 667.5 \\ \text { test statistic }= & 1.47809 \\ \text { D. F. }= & 40 \\ \text { P-VALUE }= & 0.147626 \\ \text { P-VALUE/2 }= & 0.0738131 \\ \text { SD.ERROR }= & 451.597 \end{array}\) What of the following assumptions is necessary to perform the test described above?

A) The means of the two populations of salaries are equal. B) The population of salaries for each of the two industries has an approximately normal distribution. C) The standard deviations of the two populations of salaries are both large. D) None. The Central Limit Theorem takes care of all assumptions A) The means of the two populations of salaries are equal. B) The population of salaries for each of the two industries has an approximately normal distribution. C) The standard deviations of the two populations of salaries are both large. D) None. The Central Limit Theorem takes care of all assumptions

In order to compare the means of two populations, independent random samples of 144 observations are selected from each population with the following results. $\begin{array}{ll} \text { Sample 1 } & \text { Sample 2 } \\ \bar{x}_{1}=7,123 & \bar{x}_{2}=6,957 \\ s_{1}=175 & s_{2}=225 \end{array}$ Use a 95\% confidence interval to estimate the difference between the population means $\left( \mu _ { 1 } - \mu _ { 2 } \right)$. Interpret the confidence interval.

The answer of In order to compare the means of...

Exam 8: Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses

When blood levels are low at an area hospital, a call goes out to local residents to give blood. The blood center is interested in determining which sex - males or females - is more likely to respond. Random, independent samples of 60 females and 100 males were each asked if they would be willing to give blood when called by a local hospital. A success is defined as a person who responds to the call and donates blood. The goal is to compare the percentage of the successes between the male and female responses. What type of analysis should be used?

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Question 1

Correct Answer:

Verified

C

Find $F _ { 10 } \text { where } v _ { 1 } = 20 \text { and } v _ { 2 } = 40 \text {. }$

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Question 2

Correct Answer:

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A

An inventor has developed a new spray coating that is designed to improve the wear of bicycle tires. To test the new coating, the inventor randomly selects one of the two tires on each of 50 bicycles to be coated with the new spray. The bicycle is then driven for 100 miles and the amount of the depth of the tread left on the two bicycle tires is measured (in millimeters). It is desired to determine whether the new spray coating improves the wear of the bicycle tires. The data and summary information is shown below: Bicycle Coated Tire () Non-Coated Tire (N) 1 1.452 0.785 2 1.634 0.844 \downarrow \downarrow \downarrow 50 1.211 0.954 Coated Non-Coated Difference Mean 1.38 0.85 0.53 Std. Dev. 0.12 0.11 0.06 Sample Size 50 50 50 Identify the correct null and alternative hypothesis for testing whether the new spray coating improves the mean wear of the bicycle tires (which would result in a larger amount of tread left on the tire).

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Question 3

Correct Answer:

Verified

C

Determine whether the sample sizes are large enough to conclude that the sampling distributions are approximately normal. $n _ { 1 } = 48 , n _ { 2 } = 55 ,\hat { p} _ { 1 } = .4 ,\hat { }p _ { 2 } = .7$

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Question 4

Which of the following represents the difference in two population means?

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Question 5

A government housing agency is comparing home ownership rates among several immigrant groups. In a sample of 235 families who emigrated to the U.S. from Eastern europe five years ago, 165 now own homes. In a sample of 195 families who emigrated to the U.S. from Pacific islands five years ago, 125 now own homes. Write a 95% confidence interval for the difference in home ownership rates between the two groups. Based on the confidence interval, can you conclude that there is a significant difference in home ownership rates in the two groups of immigrants?

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Question 6

In a controlled laboratory environment, a random sample of 10 adults and a random sample of 10 children were tested by a psychologist to determine the room temperature that each person finds most comfortable. The data are summarized below: Sample Mean Sample Variance Adults (1) 77. 4.5 Children (2) 74. 2.5 Suppose that the psychologist decides to construct a 99% confidence interval for the difference in mean comfortable room temperatures instead of proceeding with a test of hypothesis. The 99% confidence interval turns out to be (-2.9, 3.1). Select the correct statement.

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Question 7

University administrators are trying to decide where to build a new parking garage on campus. The state legislature has budgeted just enough money for one parking structure on campus. The administrators have determined that the parking garage will be built either by the college of engineering or by the college of business. To help make the final decision, the university has Randomly and independently asked students from each of the two colleges to estimate how long they usually take to find a parking spot on campus (in minutes). Based on their sample, the following 95% confidence interval (for µe - µb) was created - (4.20, 10.20). What conclusion can the university make about the population mean parking times based on this confidence interval?

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Question 8

The FDA is comparing the mean caffeine contents of two brands of cola. Independent random samples of 6-oz. cans of each brand were selected and the caffeine content of each can determined. the study provided the following summary information. Brand A Brand B Sample size 15 10 Mean 18 20 Variance 1.2 1.5 How many cans of each soda would need to be sampled in order to estimate the difference in the mean caffeine content to within . 5 with $95 \%$ reliability?

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Question 9

Determine whether the sample sizes are large enough to conclude that the sampling distributions are approximately normal. $n _ { 1 } = 45 , n _ { 2 } = 52 , \hat { p } _ { 1 } = .3 , \hat { p } _ { 2 } = .6$

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Question 10

Consider the following set of salary data: Men (1) Women (2) Sample Size 100 80 Mean \ 12,850 \ 13,000 Standard Deviation \ 345 \ 500 Suppose the test statistic turned out to be $z = - 1.20$ (not the correct value). Find a two-tailed $p$ -value for this test statistic.

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Question 11

In a controlled laboratory environment, a random sample of 10 adults and a random sample of 10 children were tested by a psychologist to determine the room temperature that each person finds most comfortable. The data are summarized below: Sample Mean Sample Variance Adults (1) 77. 4.5 Children (2) 74. 2.5 If the psychologist wished to test the hypothesis that children prefer warmer room temperatures than adults, which set of hypotheses would he use?

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Question 12

Identify the rejection region that should be used to test $H _ { 0 } : \sigma _ { 1 } ^ { 2 } = \sigma _ { 2 } { } ^ { 2 }$ against $H _ { \mathrm { a } } : \sigma _ { 1 } ^ { 2 } > \sigma _ { 2 } { } ^ { 2 }$ for $v _ { 1 } = 10 , v _ { 2 } = 29$ , and $\alpha = .10$ .

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Question 13

A cola manufacturer invited consumers to take a blind taste test. Consumers were asked to decide which of two sodas they preferred. The manufacturer was also interested in what factors played a role in taste preferences. Below is a printout comparing the taste preferences of men and women. HYPOTHESIS: PROP. $X =$ PROP. $Y$ SAMPLES SELECTED FROM soda(brand1, brand2) males ( sex =0, males ) ( NUMBER =115) females ( sex =1, females ) ( NUMBER =56) $X=$ males $\mathrm{Y}=$ females SAMPLE PROPORTION OF X $= 0.422018$ SAMPLE SIZE OF $X = 109$ SAMPLE PROPORTION OF $Y = 0.25$ SAMPLE SIZE OF $\mathrm { Y } = 52$ PROPORTION X - PROPORTION Y = $0.172018$ $Z = 2.11825$ Suppose the manufacturer wanted to test to determine if the males preferred its brand more than the females. Using the test statistic given, compute the appropriate p-value for the test.

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Question 14

Salary data were collected from CEOs in the consumer products industry and CEOs in the telecommunication industry. The data were analyzed using a software package in order to compare mean salaries of CEOs in the two industries. $\text { HYPOTHESIS: MEAN X = MEAN Y }$ $\text { SAMPLES SELECTED FROM SALARY }$ X= Consumer Products = Telecommunications SAMPLE MEAN OF X= 1761 SAMPLE VARIANCE OF X = 3.97555 E 6 SAMPLE SIZE OF X = 21 SAMPLE MEAN OF Y = 1093.5 SAMPLE VARIANCE OF Y = 103255 SAMPLE SIZE OF Y = 21 MEAN X - MEAN Y = 667.5 test statistic = 1.47809 D. F. = 40 P-VALUE = 0.147626 P-VALUE/2 = 0.0738131 SD.ERROR = 451.597 What of the following assumptions is necessary to perform the test described above?

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Question 15

In order to compare the means of two populations, independent random samples of 144 observations are selected from each population with the following results. Sample 1 Sample 2 =7,123 =6,957 =175 =225 Use a 95\% confidence interval to estimate the difference between the population means $\left( \mu _ { 1 } - \mu _ { 2 } \right)$ . Interpret the confidence interval.

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Question 16

Which supermarket has the lowest prices in town? All claim to be cheaper, but an independent agency recently was asked to investigate this question. The agency randomly selected 100 items common to each of two supermarkets (labeled A and B) and recorded the prices charged by each supermarket. The summary results are provided below: =2.09 =1.99 =.10 =0.22 =0.19 =.03 Assuming the data represent a matched pairs design, calculate the confidence interval for comparing mean prices using a 95% confidence level.

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Question 17

Construct a $90 \%$ confidence interval for $\left( p _ { 1 } - p _ { 2 } \right)$ when $n _ { 1 } = 400 , n _ { 2 } = 550 , \hat { p } _ { 1 } = .42$ , and $\hat { p } _ { 2 }$ $= .63$ .

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Question 18

$\text { The screens below show the results of a test of } H_{0}:\left(\mu_{1}-\mu_{2}\right)=0 \text { against } H_{a}:\left(\mu_{1}-\mu_{2}\right) \neq 0$ $\text { The screens below show the results of a test of } H_{0}:\left(\mu_{1}-\mu_{2}\right)=0 \text { against } H_{a}:\left(\mu_{1}-\mu_{2}\right) \neq 0 Comment on the validity of the results.$ Comment on the validity of the results.

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Question 19

One indication of how strong the real estate market is performing is the proportion of properties that sell in less than 30 days after being listed. Of the condominiums in a Florida beach community that sold in the first six months of 2006, 75 of the 115 sampled had been on the market less than 30 days. For the first six months of 2007, 25 of the 85 sampled had been on the market less than 30 days. Test the hypothesis that the proportion of condominiums that sold within 30 days decreased from 2006 to 2007. Use $\alpha = .01$

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Question 20

Find $F _ { 10 } \text { where } v _ { 1 } = 20 \text { and } v _ { 2 } = 40 \text {. }$

Determine whether the sample sizes are large enough to conclude that the sampling distributions are approximately normal. $n _ { 1 } = 48 , n _ { 2 } = 55 ,\hat { p} _ { 1 } = .4 ,\hat { }p _ { 2 } = .7$

Which of the following represents the difference in two population means?

Determine whether the sample sizes are large enough to conclude that the sampling distributions are approximately normal. $n _ { 1 } = 45 , n _ { 2 } = 52 , \hat { p } _ { 1 } = .3 , \hat { p } _ { 2 } = .6$

Consider the following set of salary data: Men (1) Women (2) Sample Size 100 80 Mean \ 12,850 \ 13,000 Standard Deviation \ 345 \ 500 Suppose the test statistic turned out to be $z = - 1.20$ (not the correct value). Find a two-tailed $p$ -value for this test statistic.

Identify the rejection region that should be used to test $H _ { 0 } : \sigma _ { 1 } ^ { 2 } = \sigma _ { 2 } { } ^ { 2 }$ against $H _ { \mathrm { a } } : \sigma _ { 1 } ^ { 2 } > \sigma _ { 2 } { } ^ { 2 }$ for $v _ { 1 } = 10 , v _ { 2 } = 29$ , and $\alpha = .10$ .

Construct a $90 \%$ confidence interval for $\left( p _ { 1 } - p _ { 2 } \right)$ when $n _ { 1 } = 400 , n _ { 2 } = 550 , \hat { p } _ { 1 } = .42$ , and $\hat { p } _ { 2 }$ $= .63$ .

Statistics, Data, and Statistical Thinking

Methods for Describing Sets of Data

Probability

Random Variables and Probability Distributions

Sampling Distributions

Inferences Based on a Single Sample: Estimation With Confidence Intervals

Inferences Based on a Single Sample: 355 Tests of Hypotheses

Design of Experiments and Analysis of Variance

Categorical Data Analysis

Simple Linear Regression

Multiple Regression and Model Building

Methods for Quality Improvement: Statistical Process Control Available on CD

Time Series: Descriptive Analyses, Models, and Forecasting Available on CD

Nonparametric Statistics Available on CD

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Exam 8: Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses

Find F10 where v1=20 and v2=40. F _ { 10 } \text { where } v _ { 1 } = 20 \text { and } v _ { 2 } = 40 \text {. }F10​ where v1​=20 and v2​=40.

Determine whether the sample sizes are large enough to conclude that the sampling distributions are approximately normal. n1=48,n2=55,p^1=.4,^p2=.7n _ { 1 } = 48 , n _ { 2 } = 55 ,\hat { p} _ { 1 } = .4 ,\hat { }p _ { 2 } = .7n1​=48,n2​=55,p^​1​=.4,^p2​=.7

Which of the following represents the difference in two population means?

Determine whether the sample sizes are large enough to conclude that the sampling distributions are approximately normal. n1=45,n2=52,p^1=.3,p^2=.6n _ { 1 } = 45 , n _ { 2 } = 52 , \hat { p } _ { 1 } = .3 , \hat { p } _ { 2 } = .6n1​=45,n2​=52,p^​1​=.3,p^​2​=.6

Consider the following set of salary data: Men (1) Women (2) Sample Size 100 80 Mean \ 12,850 \ 13,000 Standard Deviation \ 345 \ 500 Suppose the test statistic turned out to be z=−1.20z = - 1.20z=−1.20 (not the correct value). Find a two-tailed ppp -value for this test statistic.

Construct a 90%90 \%90% confidence interval for (p1−p2)\left( p _ { 1 } - p _ { 2 } \right)(p1​−p2​) when n1=400,n2=550,p^1=.42n _ { 1 } = 400 , n _ { 2 } = 550 , \hat { p } _ { 1 } = .42n1​=400,n2​=550,p^​1​=.42 , and p^2\hat { p } _ { 2 }p^​2​ =.63= .63=.63 .

Statistics, Data, and Statistical Thinking

Methods for Describing Sets of Data

Probability

Random Variables and Probability Distributions

Sampling Distributions

Inferences Based on a Single Sample: Estimation With Confidence Intervals

Inferences Based on a Single Sample: 355 Tests of Hypotheses

Design of Experiments and Analysis of Variance

Categorical Data Analysis

Simple Linear Regression

Multiple Regression and Model Building

Methods for Quality Improvement: Statistical Process Control Available on CD

Time Series: Descriptive Analyses, Models, and Forecasting Available on CD

Nonparametric Statistics Available on CD

Filters

Find $F _ { 10 } \text { where } v _ { 1 } = 20 \text { and } v _ { 2 } = 40 \text {. }$

Determine whether the sample sizes are large enough to conclude that the sampling distributions are approximately normal. $n _ { 1 } = 48 , n _ { 2 } = 55 ,\hat { p} _ { 1 } = .4 ,\hat { }p _ { 2 } = .7$

Determine whether the sample sizes are large enough to conclude that the sampling distributions are approximately normal. $n _ { 1 } = 45 , n _ { 2 } = 52 , \hat { p } _ { 1 } = .3 , \hat { p } _ { 2 } = .6$

Consider the following set of salary data: Men (1) Women (2) Sample Size 100 80 Mean \ 12,850 \ 13,000 Standard Deviation \ 345 \ 500 Suppose the test statistic turned out to be $z = - 1.20$ (not the correct value). Find a two-tailed $p$ -value for this test statistic.

Construct a $90 \%$ confidence interval for $\left( p _ { 1 } - p _ { 2 } \right)$ when $n _ { 1 } = 400 , n _ { 2 } = 550 , \hat { p } _ { 1 } = .42$ , and $\hat { p } _ { 2 }$ $= .63$ .