If n=41n=41, ¯xx¯=36, and s=9s=9, find the margin of error for a 80% confidence level.

Q: Use the Student's t distribution to find t, for a 0.95 confidence level when the sample is 27.

A: sample size=n=27 Degree of Freedom=n-1=27-1 =26 Confidence level is 0.95 Therefore the Level of…

Q: Out of a sample of 600 adults aged 18 to 30, 162 still lived with their parents. Based on this,…

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Q: If n = 10, ¯xx¯ = 43, and s = 8, construct a confidence interval at a 90% confidence level. Assume…

A: Since population standard deviation is unknown, Use t-distribution to find t-critical value. Find…

Q: If n=30, (x-bar)=38, and s=8, construct a confidence interval at a 90% confidence level. Assume the…

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Q: If n=28, (x-bar)=50, and s=10, construct a confidence interval at a 99% confidence level. Assume the…

A: Sample size n=28Sample mean=50Standard deviation =10

Q: If n = 108, a = 29 and o = 2.6, construct a 91% confidence interval. Give your answers to three…

A: Since we know population standard deviation, use z-distribution to find z-critical value.…

Q: Construct the indicated confidence interval for the difference between population proportions p1 -…

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Q: a 95% confidence interval estimate for population mean is determined to be 75.38 to 86.52. if the…

A: We have  to find the confidence level is reduced to 90%, the confidence interval will....

Q: If n=10, ī(x-bar)=37, and s=19, construct a confidence interval at a 95% confidence level. Assume…

A: From the provided information, Sample size (n) = 10 Sample mean (x̄) = 37 Standard deviation (s) =…

Q: If n=31, (x-bar)=49, and s=3, construct a confidence interval at a 80% confidence level. Assume the…

A: Given, Sample size = 31 Sample mean = 49 Sample standard deviation = 3   Since the population…

Q: If n = 22, ¯xx¯ = 36, and s = 14, construct a confidence interval at a 98% confidence level. Assume…

A: From the provided information, Sample size (n) = 22 Sample mean (x̄) = 36 Standard deviation (s) =…

Q: If n = 590 and ˆp=0.48, find the margin of error at a 99% confidence level E=

A: If n = 590 and ˆp=0.48, find the margin of error at a 99% confidence level

Q: If n = 28, ¯xx¯ = 49, and s = 19, construct a confidence interval at a 90% confidence level. Assume…

A: Given information- Sample size, n = 28 Sample mean, x-bar = 49 Sample standard deviation, s = 19…

Q: if E=0.07 and population standard deviation=0.8 then find the sample size for 99% confidence…

A: Margin of error(E)=0.07 population standard deviation(σ)=0.8 confidence level=99%

Q: construct a confidence interval at a 98% confidence level.

A: Given : n=29 , X-bar=39  , s=11  , α=0.02 Since , the population standard deviation (σ) is unknown.…

Q: Given a sample size of 400 with a mean of 80 and the population standard deviation is found to be 35…

A: We have given that,  Sample mean (x̄) = 80, population standard deviation (σ) = 35 and sample size…

Q: If n = 16, 7 = 43, and s = 17, construct a confidence interval at a 98% confidence level. Assume the…

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Q: If n=13, (x-bar)=48, and s=12, construct a confidence interval at a 95% confidence level. Assume the…

A: Mean = 48Standard deviation = 12Sample size = 13

Q: If n=12, (x-bar)=33, and s=16, construct a confidence interval at a 80% confidence level. Assume the…

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Q: If n = 25, ¯x = 50, and s = 19, construct a confidence interval at a 99% confidence level. Assume…

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Q: If n = 15, ¯x = 50, and s = 9, construct a confidence interval at a 90% confidence level. Assume the…

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Q: one with a 95% confidence level and the other with a 99.7% confidence level. Which interval is…

A: Confidence interval is probability, which refers that population parameter will fall within set of…

Q: Construct the indicated confidence interval for the difference between population proportions p1-p2.…

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Q: If n=12, x=47, and s=7, construct a confidence interval at a 98% confidence level. Assume the data…

A: According to the provided information,

Q: If n=27, x=48, and s-6, construct a confidence interval at a 80% confidence level. Assume the data…

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Q: If n = 17, ¯xx¯ = 33, and s = 6, construct a confidence interval at a 99% confidence level. Assume…

A: We have, If n = 17, x¯ = 33, and s = 6, alpha=1-0.99= 0.01 Therefore  Degrees of freedom= n-1=16…

Q: If n = 18, = 46, and s = 16, construct a confidence interval at a 90% confidence level. Assume the…

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Q: 13, ¤(x-bar)=39, and s=16, construct a confidence interval at a 90% confidence level. Assume the…

A: From the given information, n=13 x-bar=39 s=16 Confidence level=90% Significance level=10%…

Q: If n 28, = 37, and s = 6.2, construct a confidence interval at a 97% confidence level. Assume the…

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

Q: If n = 41, ã=30, and s = 6, construct a confidence interval at a 80% confidence level.

A: It is given that Sample size n = 41 Sample mean = 30 Sample SD s = 6 The critical value of t at df =…

Q: If n = 180 and X = 144, construct a 99% confidence interval for the population proportion, p. Give…

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Q: If n = 380 and X = 304, construct a 95% confidence interval for the population proportion, p. Give…

A: Given: n = 380 x = 304

Q: If n=14, (x-bar)=46, and s=18, construct a confidence interval at a 80% confidence level. Assume the…

A: Since population standard deviation is unknown, Use t-distribution to find t-critical value. Find…

Q: If n = 15, ¯xx¯ = 43, and s = 11, construct a confidence interval at a 98% confidence level. Assume…

A: Given that, The degree of freedom is, Critical values: The t-critical values for a two-tailed…

Q: IF n=23, x-bar = 39, and s=3, find the margin of error at a 80% confidence level. Give answer to two…

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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…

Q: identify the t-score for a 90% confidence interval if the sample size is 31.        t=

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Q: FIND Za2 FOR THE 99.50 CONFIDENCE INTERVAL

A: Find α/2Level of Confidence = 99.5%α = 100% - (Level of Confidence) = 0.5%α/2 = 0.25% = 0.0025

Q: In a survey conducted to 415 people it was found that they had walked 5000 meters on average for the…

A: Sample size (n) = 415Sample mean (x̅) = 5000Standard deviations (s) = 900 Confidence level (c) =…