Suppose that a random sample of 15 recently sold houses in a certain city has a mean sales price of $300,000, with a standard deviation of $11,000. Under the assumption that house prices are normally distributed, find a 95% confidence interval for the mean sales price of all houses in this city. Give the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to the nearest whole number. (If necessary, consult a list of formulas.) Lower limit: s Upper limit: s X

Q: e: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the…

A: Sample size n =47 Sample mean=109 Population standard deviation =32

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A: Given data,x=208n=260p^=xn=208260=45=0.80

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A:

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A: Given Data: Confidence level - 90%: The expression for the confidence level is,

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A: The sample size is 1060 and number of success is 169.

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A: Let X: a particular fruit's weightsX~N(μ,σ)where mean(μ)=657 grams  standard deviation(σ)=38 grams…

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A: Solution: From the given information, X=144, n=180 and the confidence level is 0.90. The sample…

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A:

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A: We have given that,  Sample size (n) = 56, population standard deviation (σ) = 35 and sample mean…

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A: According to the sum, A particular fruit's weights are normally distributed, with a mean of 596…

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A: The 95% confidence interval is (4,6).

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A: sample size(n)=48 population standard deviation(σ)=25 Mean(x)=105

Q: If n = 330 and X = 264, construct a 95% confidence interval for the population proportion, p.

A: Ans#- From the provided information, Confidence level = 95% Sample size (n) = 330, x = 264 Sample…

Q: The 90% confidence interval is from to. (Round to three decimal places as needed.)

A: A poll of 1030 adults in a country was conducted by telephone. Of those surveyed, 139 had…

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A: Given Mean=724 Standard deviation=9 n=18

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A: n = 43 standard deviation = 27 mean = 104 95% confidence interval for mean is given by  x±zα2*σn 104…

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A: It is given that, Sample mean = 104. Population standard deviation, sigma is 31. The sample size, n…

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A: Sample size n =49 Sample mean=103 Population standard deviation =25

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A:

Q: lculating the confidence interval.

A: x Freq 1 3 2 5 3 19 4 7 5 6 Total 40

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A: Solution: The following information is given to us: s=6.2 n =28 A 97% confidence interval is…

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A: Given: Suppose that the responses for minutes spent studying per week range from 0 to 480 minutes.…

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A: Normal distribution is a continuous probability distribution which has many real life applications.…

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A: Standard Deviation = σ = 31 Margin of Error = E = 3 Confidence Level = 99 Significance Level =…

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A: Sample size, Population standard deiation, Sample mean,

Q: A statistics practitioner took a random sample of 47 observations from a population whose standard…

A: Given that  Sample size n =47 Sample mean =97 Population standard deviation =32

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A: sample size(n)=59 Population standard deviation(σ)=32 Mean(x)=95

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A: Here, the sample proportion is =7911495=0.5291 Now, for 90% confidence z=1.645

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Q: If n=21, (x-bar)=37, and s=3, construct a confidence interval at a 90% confidence level. Assume the…

A: Given Information: Sample size n=21 Sample mean x¯=37 Sample standard deviation s=3 Confidence level…

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A: We have given that,  sample mean (x̄) = 109, standard deviation (σ) = 30 and sample size (n) = 43

Q: Mason earned a score of 226 on Exam A that had a mean of 250 and a standard deviation of 40. He is…

A: First find the standardised Z score and the make equivalency

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A: Given information Sample size (n) = 49 Sample mean x̅ = 55.1 Standard deviation (s) = 4.8…