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Exam 9: Statistical Inferences Based on Two Samples

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Question 31

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Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients.In 1996,the average LOS for non-heart patient was 4.6 days.A random sample of 20 hospitals had a mean LOS for non-heart patients in 2000 of 3.8 days and a standard deviation of 1.2 days.Assume LOS is normally distributed.How large a sample of hospitals would we need to be 99% confident that the sample mean is within 0.5 days of the population mean?

A) 3
B) 7
C) 32
D) 48
E) 96

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Question 32

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In general,replacing a z with a t when using a preliminary sample to obtain an estimate of the variance is a _____ decision.

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Question 33

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The provincial transportation department is studying traffic patterns on one of the busiest highways in the province.As part of the study,the department needs to estimate the average number of vehicles that pass an intersection each day.A random sample of 64 days gives us a sample mean of 14,205 cars.Historically,the population standard deviation has been approximately 1,010 cars.What is the 92% confidence interval estimate of $\mu$ ,the mean number of cars passing the intersection?

The 95% confidence interval for the average weight of a product is from 72.23kg to 77.77kg.Interpret this confidence interval.

A company is interested in estimating $\mu$ , the mean number of days of sick leave taken by its employees. The firm's statistician randomly selects 100 personnel files and notes the number of sick days taken by each employee. The sample mean is 12.2 days. From historical records, the population standard deviation is about 10 days. -How many personnel files would the director have to select in order to estimate $\mu$ To within 2 days with 99% confidence?

In a survey of 400 people,60 percent favour new zoning laws.Find a 95 percent confidence interval for the true proportion favouring new laws.

Sandwiches sold at a local grocery store claim to have 6 grams of fat per sandwich.Ten sandwiches are brought with results of a mean of 5.93g of fat and a standard deviation of 0.13g.If fat content is normally distributed,how large a sample of sandwiches would we need to be 95% confident that the sample mean is within 0.03g of the population mean?

When computed based on the same sample,the 99% confidence interval for a population mean $\mu$ will be narrower than the 95% confidence interval for $\mu$ .

As the margin of error decreases,the width of the confidence interval _______________.

A granola bar manufacturer states that their granola bars contain 28 grams of carbohydrates per bar.A sample of five granola bars is tested and a mean of 28.2g is obtained with a standard deviation of 2.7g.If carbohydrate content is normally distributed,the 95% confidence interval for the mean carbohydrate content of all such granola bars made by this company would be _____.

A random sample of size 10 is taken from a population assumed to be normal,and $\bar { x }$ = 1.2 and s = 0.6.Calculate a 90 percent confidence interval for $\mu$ .

The Ministry of Agriculture tested 105 fuel samples for accuracy of the reported octane level.For premium grade,14 out of 105 samples failed (they didn't meet specifications) .The 99% confidence interval for the true population proportion of premium grade fuel-quality failures would be _____.

When the level of confidence and sample proportion $\hat { p }$ Remain the same,a confidence interval for a population proportion p based on a sample of n = 100 will be ______________ a confidence interval for p based on a sample of n = 400.

Consider a normally distributed population.Which two of the following actions will result in the greatest increase in the width of a confidence interval for the mean? (i) increasing the confidence level (ii) decreasing the confidence level (iii) increasing the sample size (iv) decreasing the sample size (v) decreasing the standard deviation

A test of spelling ability is given to a random sample of 10 students before and after they completed a spelling course.The mean score before the course was 119.60 and after the course the mean score was 130.80.The standard deviation of the difference was 16.061.Assume that the difference in scores is normally distributed.Calculate a 99% confidence interval for the mean change in score.

Suppose that a realtor is interested in comparing the price of a starter home in two small cities.The realtor conducts a small survey in the two cities looking at the price of starter homes.Assume equal population variances and normally distributed populations. $\begin{array} { l r r } & { \text { City A } } & \text { City B } \\\text { Sample Mean } & 86,900 & 84,000 \\\text { Sample Std Dev } & 2300 & 1750 \\\text { Sample Size } & 9 & 7\end{array}$ -Calculate the 95% confidence interval for the difference in mean price.

When a confidence interval for a population proportion is constructed for a sample size of 30 and the value of $\hat { p }$ Is 0.4,the confidence interval is based on the ____ distribution.

A sample set of weights in kilograms are 1.01,.95,1.03,1.04,.97,.97,.99,1.01,and 1.03.Assume the population of weights are normally distributed.Find a 99 percent confidence interval for the mean population weight.

When constructing a confidence interval for a population mean,if the population is normally distributed with known standard deviation and the sample size is n = 5,then the distribution of $\bar { x }$ Is based on _____ distribution.

If we wish to estimate a population proportion to within 5% with 95% confidence,the required sample size is 385.