Exam 9: Statistical Inferences Based on Two Samples

Find a 95% confidence interval for the difference between the proportions of people who prefer cola versus root beer where $\hat { p } _ { \text {cola } }$ = .21, $\hat { p } _ { \text {root beer } }$ = .12,n_cola = 200,n_{root beer} = 150.

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Question 161

Assuming the same confidence level $1 - \alpha$ ,as the sample size increases,the value of $t _ { \alpha / 2 }$ approaches the value of $z _ { \alpha / 2 }$ .

(True/False)

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Question 162

A manager is interested in the amount of time employees spend each day playing video games at work.The manager takes a random sample of 18 employees who showed a mean time of 55 minutes per day with a standard deviation of 5 minutes.Assuming that these times follow a normal distribution,compute a 90% confidence interval for the average amount of time employees spend playing video games each day at work.

(Multiple Choice)

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Question 163

In 2007,13,000 internet users were surveyed and asked about their willingness to pay fees for access to websites.Of these,2,938 were definitely not willing to pay such fees.How large a sample is necessary to estimate the proportion of interest to within 2% in a 95% confidence interval?

(Multiple Choice)

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Question 164

An injector molder produces plastic pens.The process is designed to produce pens with a mean weight of 0.250 grams.To investigate whether the injection molder is operating satisfactorily,40 pens were randomly sample with a mean of 0.2525g.The population standard deviation is believed to be approximately 0.0022g.Find a 99% confidence interval for the population mean.

(Multiple Choice)

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Question 165

Given the following test scores,find a 95 percent confidence interval for the population mean: 148,154,158,160,161,162,166,170,182,195,236.Assume population normality.

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Question 166

In a study of 265 subjects,the average score on the examination was 63.8 and s = 3.08.What is a 95 percent confidence for $\mu$ ? Assume scores are normally distributed.

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Question 167

A coffee shop franchise owner is looking at two possible locations for a new shop. To help make a decision the owner looks at the number of pedestrians that go by each of the two locations in one-hour segments. At location A, counts are taken for 35 one-hour units and with a mean number of pedestrians of 421 and a sample standard deviation of 122. At the second location (B), counts are taken for 50 one hour units with a mean number of pedestrians of 247 and a sample standard deviation of 85. Assume the two populations variances are not known but are equal, and that the number of pedestrians is approximately normal. -Calculate a 95% confidence interval for the difference in pedestrian traffic at the two locations.

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Question 168

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Find a 95% confidence interval for the difference between the proportions of people who prefer cola versus root beer where $\hat { p } _ { \text {cola } }$ = .21, $\hat { p } _ { \text {root beer } }$ = .12,n_cola = 200,n_{root beer} = 150.

Assuming the same confidence level $1 - \alpha$ ,as the sample size increases,the value of $t _ { \alpha / 2 }$ approaches the value of $z _ { \alpha / 2 }$ .

In 2007,13,000 internet users were surveyed and asked about their willingness to pay fees for access to websites.Of these,2,938 were definitely not willing to pay such fees.How large a sample is necessary to estimate the proportion of interest to within 2% in a 95% confidence interval?

Given the following test scores,find a 95 percent confidence interval for the population mean: 148,154,158,160,161,162,166,170,182,195,236.Assume population normality.

In a study of 265 subjects,the average score on the examination was 63.8 and s = 3.08.What is a 95 percent confidence for $\mu$ ? Assume scores are normally distributed.

An Introduction to Business Statistics

Descriptive Statistics

Probability

Discrete Random Variables

Continuous Random Variables

Sampling Distributions

Confidence Intervals

Hypothesis Testing

Experimental Design and Analysis of Variance

Correlation Coefficient and Simple Linear Regression Analysis

Multiple Regression and Model Building

Nonparametric Methods

Chi-Square Tests

Decision Theory

Time Series Forecasting

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Exam 9: Statistical Inferences Based on Two Samples

Find a 95% confidence interval for the difference between the proportions of people who prefer cola versus root beer where p^cola \hat { p } _ { \text {cola } }p^​cola ​ = .21, p^root beer \hat { p } _ { \text {root beer } }p^​root beer ​ = .12,ncola = 200,nroot beer = 150.

Assuming the same confidence level 1−α1 - \alpha1−α ,as the sample size increases,the value of tα/2t _ { \alpha / 2 }tα/2​ approaches the value of zα/2z _ { \alpha / 2 }zα/2​ .

In 2007,13,000 internet users were surveyed and asked about their willingness to pay fees for access to websites.Of these,2,938 were definitely not willing to pay such fees.How large a sample is necessary to estimate the proportion of interest to within 2% in a 95% confidence interval?

Given the following test scores,find a 95 percent confidence interval for the population mean: 148,154,158,160,161,162,166,170,182,195,236.Assume population normality.

In a study of 265 subjects,the average score on the examination was 63.8 and s = 3.08.What is a 95 percent confidence for μ\muμ ? Assume scores are normally distributed.

An Introduction to Business Statistics

Descriptive Statistics

Probability

Discrete Random Variables

Continuous Random Variables

Sampling Distributions

Confidence Intervals

Hypothesis Testing

Experimental Design and Analysis of Variance

Correlation Coefficient and Simple Linear Regression Analysis

Multiple Regression and Model Building

Nonparametric Methods

Chi-Square Tests

Decision Theory

Time Series Forecasting

Filters

Find a 95% confidence interval for the difference between the proportions of people who prefer cola versus root beer where $\hat { p } _ { \text {cola } }$ = .21, $\hat { p } _ { \text {root beer } }$ = .12,n_cola = 200,n_{root beer} = 150.

Assuming the same confidence level $1 - \alpha$ ,as the sample size increases,the value of $t _ { \alpha / 2 }$ approaches the value of $z _ { \alpha / 2 }$ .

In a study of 265 subjects,the average score on the examination was 63.8 and s = 3.08.What is a 95 percent confidence for $\mu$ ? Assume scores are normally distributed.