Example 7. In a random selection of 64 of 600 road crossing in a town, the mean number of automobile accidents per year was found to 4.2 and the sample s.d was 0.8. Construct a 95% confidence interval for the mean number of automobile accidents per crossing per year.
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A: n=sample size= 36 X¯=sample mean=83σ=sample standard deviation=5 So the point estimate=X=83 Since…
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A: *Answer:
Q: 2. A study reported in The Journal of Consumer Research dealt with the influence of touch on…
A: Given that,Females diners who were touched:The sample size is 35.The sample mean is 3.84.The…
Q: Suppose you observe 77 data points from a distribution with standard deviation 8.4, and you…
A: Given that, σ=8.4,n=77 Critical values: By using the z-table, the critical value at 92% confidence…
Q: 3. A study is conducted on people with glaucoma with higher-than-average blood pressure. The study…
A: Given that, A study is conducted on people with glaucoma with higher-than-average blood pressure.…
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Q: From the normally distributed systolic blood pressures of a given patient population with a standard…
A: Sample mean = x̅ = 120 Sample size = n = 9 Population S.D = σ = 24 Standard Error = σ/√n =…
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Q: A recent survey conducted by the National Telecommunications Commission (NTC) aims to determine the…
A: We have to find confidence interval...
Q: In a transportation study of 27 large cities it was found that the number of motor vehicles per…
A: Given : Sample size (n) = 27 Sample mean (X¯) = one.34 i.e 1.34 Sample SD (S) = 0.46 Confidence…
Q: 5. A researcher surveyed of 300 young adults ages 18-29 found that on average, those surveyed sent…
A: 5. Given that, n=300x¯=67.7s=23.6 Confidence level=90% Significance level=1-0.90-0.10…
Q: what sample size should be used to obtain a 99% confidence interval for the mean with +/-15 if the…
A: The margin of error is 15, the standard deviation is 48 and the confidence level is 99%.
Q: suppose an individual was studying temperature's effect on crime rate. The researcher looked at 6…
A: Solution:The sample mean is obtained below:From the given information, number of observations is 6…
Q: The mayor is interested in finding a 95% confidence interval for the mean number of pounds of trash…
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Q: According to the American Medical Association, the average annual of radiologists in the USA are…
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Q: A survey of 45,535 adults showed that 20,901 were very concerned abut climate change. What is the…
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Q: Noise levels in a hospital were measured at 81 different locations. The average level was found to…
A: It is given that, Sample mean = 61.2 decibels. Sample standard deviation, s is 7.9 decibels. The…
Q: In a Pew Research poll, 287 out of 522 randomly selected U.S. men were able to identify Egypt when…
A: Given that,287 out of 522 randomly selected U.S. men were able to identify Egypt.Whereas, 233 out of…
Q: 4. Assume that 66 % of single women and 81.9% of single men own cars. Also assume that these…
A: From the provided information,
Q: In a random sample of 42 Pa high schools it is found that the average number of students is 279.…
A: Given: n=42, x=279 and σ2=206. The value of z at 93% confidence level is 1.81.
Q: 2. In a recent study of 35 college students, the mean number of hours per week that they played…
A: According to the given information in this question We need to find the 95% confidence interval
Q: A survey of 45,535 adults showed that 20,901 were very concerned abut climate change. What is the…
A: It is given asFavourable cases, x = 20901Sample size, n = 45535Confidence level = 99%
Q: 5. Assume that in a sample 450 term papers 47% of term papers were longer than two pages. Provide a…
A: Givensample size(n)=450sample proportion(p^)=0.47confidence level=98%
Q: In 2010, the proportion of Texans without health insurance was 0.11. To update this information a…
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Q: the following random sample was selected from a normal distribution: 11, 16, 3, 13, 18, 18, 16, 19,…
A: Given data is11, 16, 3, 13, 18, 18, 16, 19, 17, 11sample size(n)=10
Q: a simlpe random sample of 8 college freshmen were asked how many hours of sleep they typically got…
A: Box plot can be used to find the outlier. Box plot: Software procedure to draw box-whisker plot…
Q: At eight different locations, paint used to mark crosswalks deteriorated after being driven over by…
A: Solution: Let X be the number of cars that would cause deterioration. From the given information,…
Q: 3. The sample (1,2,3) is randomly selected from a normal population. Construct a 98% confidence…
A: Given Sample size, n=3 Sample = 1,2,3 Confidence interval =98% α=1-0.98α=0.02zα2=2.326
Q: A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each…
A: When the given sample size is less than 30 we use t-test as a test statistic. Using t-test we can…
Q: I asked 45 randomly selected WSU college students how many US states they had visited. The students'…
A: The sample size is 45, sample mean is 3.8 and standard deviation is 1.5. The degrees of freedom is,…
Q: 5.) A recent study of 28 fantastic employees at KHS showed that their mean distance traveled from…
A: Solution : Given : Let, X : Distance travelled by employees from home to work n = 28…
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Q: 12) A state fish hatchery raises trout for stocking streams and lakes. The size of the fish at…
A: GivenClaim : If fish being released have population mean less than 14 ounces. Null hypothesis :…
Q: In a randomly selected group of 650 automobile deaths, 180 were alcohol-related. Construct a 95%…
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Q: 3. A random of families in Solano St. Intramuros, resulted in the following family sizes: 5, 6, 7,…
A: Given data observations : 5,6,7,4,3,8,2,9 we have to construct 95% ci for mean. sample size (n) =…
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Q: Suppose you observe 62 data points from a distribution with standard deviation 2.1, and you…
A: Given, n=62 data points standard deviation SD= 2.1, calculated a 92% two-sided confidence interval…
Q: 4. A sociologist discovered that an average of 7.2 jobs was held by 50 retired men during the course…
A: Sample mean x̄=7.2,n=sample size=50, population standard deviations σ=2.1 Construct 95% CI for…
Q: In a simple random sample of 144 households in a county in Virginia, the average number of children…
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Q: 10)The superintendent of a golf course believes that the mean score for professional golfers on his…
A: We have to find confidence interval for population mean. Since population standard deviation () now…
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- A random survey of 5000 businesses was conducted by Statistics Canada. They reported an average of 8.98 bankruptcies per 1000 businesses from 1999 to 2005 with a standard ERROR of 1.25. Construct a 95% confidence interval for the mean number of bankruptcies per 1000 businesses between 1999 and 2005.2. A sample of six college basketball player had an average weight of 217 pounds with a sample standard deviation of 10 pounds. Find the 95% confidence interval of the true mean weight of all basketball player.12) A state fish hatchery raises trout for stocking streams and lakes. The size of the fish at release time can be controlled to a fair degree by varying the rate of feeding. The target is a mean of 10 ounces; if the fish are too small, those who catch the fish aren’t happy. A random sample of 75 fish were weighed at time of release and it was determined that the mean was 9.66 ounces with a standard deviation of 0.86 ounces. Test to determine if the fish being released have a population mean less than 10 ounces at the 0.01 significance level.
- Suppose you work for the Department of Environment and National Resources (DENR). You want to estimate, with 95% confidence, the mean length of all tilapia in a fish hatchery pond. You take a random sample of 10 tilapia and determine the average length is 10.5 inches and the sample standard deviation is 2.9 inches. Construct a confidence interval for this sample and interpret the result.56.A random sample of 40 companies with assets over $10 million was surveyed and asked to indicate their industry and annual computer technology expense. The ANOVA comparing the average computer technology expense among three industries rejected the null hypothesis. The Mean Square Error (MSE) was 195. The following table summarized the results: Based on the comparison between the mean annual computer technology expense for companies in the tax service and food service industries, the 95% confidence interval shows an interval of -5.85 to 14.85 for the difference. This result indicates that _______________. A.There is no significant difference between the two industry technology expenses B.The interval contains a difference of 20.7 C.Companies in the food service industry spend significantly more than companies in the tax service industry D.Companies in the food service industry spend significantly less than companies in the tax service industrya group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. what is the 90% confidence interval for the mean score, u, of all students taking the test?
- 4. A fire extinguisher salesman would like to estimate the mean number of home fires started by candles each year. He obtained fire data for seven randomly selected years from the National Fire Protection Agency. His sample produced a mean of 7041 fires with a standard deviation of 1600. Create the 90% confidence interval of interest.Determine the t - value to be used to construct a 90% confidence interval for the population mean based on a sample 14.Suppose that based on a random sample, a 95% confidence interval for the mean hours slept (per day) among graduate students was found to be (6.5,6.9). what is the margin of error of this confidence interval?