Stats Online Ch. 14
In a national health survey, the total cholesterol of 2075 adults in a particular country averaged 185.2 mg/dL with a standard deviation of 41.5 mg/dL.
a) Can the Central Limit Theorem be applied to describe the distribution of the cholesterol measurements? Why or why not? No, because the Central Limit Theorem is about population parameters and not individual measurements. b) Can the Central Limit theorem be used to describe the sampling distribution model for the sample mean of all adults in the country? Why or why not? Yes, if it is assumed that these adults are a representative, random sample of all adults in this country. c) Sketch and clearly label the sampling model of the mean cholesterol levels of samples of size 2075 based on the 68-95-99.7 Rule. 182.5 goes in the middle. Take 182.5-.9 =184.3 - .9 and so on to the left. Take 182.5 - .9 and that answer minus .9 and so on to the right.
On a final project in an introductory statistics class, a student reports a 95% confidence interval for the average cost of a haircut to be ($5.50,$65.00).
There is 95% confidence that the population mean is between these two numbers.
Describe how the shape, center, and spread of t-models change as the number of degrees of freedom increases.
Shape becomes closer to Normal, center does not change, spread becomes narrower.
A sample of 20 CEOs shows total annual compensations ranging from a minimum of $0.1 to $65.97 million. The average for these 20 CEOs is $17.463 million. The histogram is shown to the right. Based on these data, a computer program found that a 95% confidence interval for the mean annual compensation of all CEOs is (negative 6.29,41.21) $M. Why should you be hesitant to trust this confidence interval?
The assumptions and conditions for a t-interval are not met. The distribution has a large outlier that is pulling the mean higher.
A nutrition lab tested 40 hot dogs to see if their mean sodium content was less than the 325-mg upper limit set by regulations for "reduced sodium" franks. A 90% confidence interval estimated the mean sodium content for this kind of hot dog at 317.2 mg to 326.8 mg. Given this, what would you tell the lab about whether the hot dogs satisfy the regulation?
The confidence interval contains the 325-mg limit. Therefore, the lab should conclude that the hot dogs do not satisfy the regulation.
A biology class conducts a bird count every week during the semester. Using the number of species counted each week, a student finds a 95% confidence interval of (16.34,18.69) for the mean number of species counted. Knowing that species have to be whole numbers, the student reports that 95% of the bird counts saw 16, 17, or 18 species. Comment on the student's report.
The confidence interval only estimates the population mean species counted. The population mean number of species counted does not have to be a whole number.
Which of the following is not a condition to check to estimate the mean of a population?
There are at least 10 successes and 10 failures in the sample.
Livestock are given a special feed supplement to see if it will promote weight gain. Researchers report that the 77 cows studied gained an average of 56 pounds, and that a 95% confidence interval for the mean weight gain this supplement produces has a margin of error of plus or minus11 pounds. Some students wrote the following conclusions. Did anyone interpret the interval correctly? Explain any misinterpretations.
a) 95% of the cows studied gained between 45 and 67 pounds. The interpretation is incorrect. The confidence interval is for the population mean, not the individual cows in the study. b) One is 95% sure that a cow fed this supplement will gain between 45 and 67 pounds. The interpretation is incorrect. The confidence interval is not for individual cows. c) One is 95% sure that the average weight gain among the cows in this study was between 45 and 67 pounds. The interpretation is incorrect. One knows the average weight gain in this study was 56 pounds. d) The average weight gain of cows fed this supplement will be between 45 and 67 pounds 95% of the time. The interpretation is incorrect. The average weight gain of all cows does not vary. It is a constant value which the confidence interval estimates. e) If this supplement is tested on another sample of cows, there is a 95% chance that their average weight gain will be between 45 and 67 pounds. The interpretation is incorrect. There is a 95% chance that another sample will have its average weight gain within two standard deviations of the true mean.
A company announced a "1000 Chips Trial," claiming that every 18-ounce bag of its cookies contained at least 1000 chocolate chips. Students purchased random bags of cookies from different stores and counted the number of chips in each bag. Some of the data is shown below. 1019 1144 1179 1292 1291 1288 1226 1374 1317 1440
a) Check the assumptions and conditions for inference. The data are from a random sample, so one can assume that they are independent. Evaluate the histogram. The histogram is roughly unimodal and symmetric, with no outliers, so one can assume that the data come from a population that follows a Normal model. b)What does this evidence say about the company's claim? Because the confidence interval is entirely above 1000, the mean number of chips per bag is likely more than 1000. However, the Normal model predicts that a small amount of individual bags will have fewer than 1000 chips.
A grocery store's receipts show that Sunday customer purchases have a skewed distribution with a mean of $33 and a standard deviation of $23. Suppose the store had 293 customers this Sunday.
a) Estimate the probability that the store's revenues were at least $10,500. The probability is $23/ square root 293 = $1.343674 10,500/293=$35.84 $35.84-$33/$1.343674=2.11 look up the z probability table to fine 2.11 = 0.0174 http://pages.stat.wisc.edu/~ifischer/Statistical_Tables/Z-distribution.pdf z=1.28 -1.28(1.343674)+33=$31.28 31.28(293)=$9165.04
A researcher measured the body temperatures of a randomly selected group of adults. He wishes to estimate the average temperature among the adult population. Summaries of the data he collected are presented in the table below. Summary Count 50 Mean 98.727 Median 98.000 MidRange 98.600 StdDev 0.8002 Range 2.800 IntQRange 1.050 Temperatures are the numbers in the summary.
a) Would a 90% confidence interval be wider or narrower than the 98% confidence interval? The 90% confidence interval would be narrower because wider intervals are more likely to contain a population parameter. b) What are the advantages and disadvantages of the 98% confidence interval? The 98% confidence interval has a greater chance of containing the true mean body temperature of adults than the 90% confidence interval, but the 98% confidence interval is less precise (wider) than the 90% confidence interval. c) If further research is conducted, this time using a sample of 500 adults, how is the 98% confidence interval likely to change? Explain. The confidence interval would likely be narrower because a larger sample would cause the standard error to be smaller.
A medical researcher measured the pulse rates (beats per minute) of a sample of randomly selected adults and found the following Student's t-based confidence interval: With 95.00% Confidence, 67.155557<u(Pulse)<70.986745
a) Explain carefully what the software output means. We are 95% confident the interval 67.2 to 71.0 beats per minute contains the true mean heart rate. b) What's the margin of error for this interval? 67.16-70.99=-3.83/2=1.9 c) If the researcher had calculated a 99% confidence interval, would the margin of error be larger or smaller? The margin of error would have been larger.
How far do the golfers at country clubs drive a ball? A histogram of the average driving distances of the 202 leading golfers at country clubs in 2002 is shown to the right, along with summary statistics. count 202 Mean 285.8 yd StdDev 8.3 yd
a) Find a 95% confidence interval for the mean drive distance. y = 285.8 s= 8.3 n=202 t=1.972 285.8-1.972/8.3 square root 202= 284.6 285.8+1.9728.3 square root 202=287.0 (284.6,287.0) b) Interpreting this interval raises some problems. Discuss. The sample is not random because only the top golfers were chosen. c) The data are the mean driving distance for each golfer. Is that a concern in interpreting the interval? Yes, the mean driving distances are less variable than the data on which they are based. This can result in an inaccurate interval.
The housing market has recovered slowly from the economic crisis of 2008. Recently, in one large community, realtors randomly sampled 32 bids from potential buyers to estimate the average loss in home value. The sample showed the average loss was $9332 with a standard deviation of $500. Suppose a 95% confidence interval to estimate the average loss in home value is found.
a) Suppose the standard deviation of the losses had been $1500 instead of $500. What would the larger standard deviation do to the width of the confidence interval (assuming the same level of confidence)? A larger standard deviation increases the width of the confidence interval. b) A classmate suggests that the margin of error in the interval could be reduced if the confidence level were changed to 90% instead of 95%. Do you agree with this statement? Why or why not? Yes, because a lower confidence interval can make the confidence interval narrower. c) Instead of changing the level of confidence, would it be more statistically appropriate to draw a bigger sample? A larger sample size would reduce the standard error of the mean and make the confidence interval narrower. It would be more statistically appropriate.
A standardized test for graduate school admission has a mean score of 152 with a standard deviation of 8 and a unimodal, symmetric distribution of scores. A test preparation organization teaches small classes of 9 students at a time. A larger organization teaches classes of 36 at a time. Both organizations publish the mean scores of all their classes.
a) What would you expect the sampling distribution of mean class scores to be for each organization? Both distributions will be approximately Normal with a mean of 152. b) If either organization has a graduating class with a mean score of 160, they'll take out a full-page ad in the local school paper to advertise. Which organization is more likely to have that success? Explain. 160-152/8/Square root 9 160-152/8/3 160-152/3=3.00 160-152/8/square root 36=6.00 The smaller organization c) Both organizations advertise that if any class has an average score below 145, they'll pay for everyone to retake the test. Which organization is at greater risk to have to pay? 145-152/8 square root 9 = -2.63 145-152/8 square root 36 = -5.25 The smaller organization
