A statistics instructor randomly selected four bags of oranges, each bag labeled 10 pounds, and weighed the bags. They weighed 9.5, 9.7,9.5, and 9.5 pounds. O Assume that the distribution of weights is Normal. Find a 95% confidence interval for the mean weight of all bags of oranges. Use technology for your calculations Answer parts a and b below. a. Choose the correct interpretation of the confidence interval below and, if necessary, fil in the answer boxes to complete your choice. OA We are 95% confident the population mean is between and OB. We are 95% confident that the sample mean is between and OC. There is a 95% chance that al intervals will be between and D. The requirements for constructing a confidence interval are not satisfied. (Type integers or decimals rounded to the nearest thousandth as needed. Use ascending order) nti

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…

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Q: If n- 22, - 44, and s 12, construct a confidence interval at a 95% confidence level. Assume the data…

A: Solution We solve it by using excel

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

A:

Q: C. If one is constructing a confidence interval for the mean and the level is 90%, then each tail…

A: c) The confidence level is 0.90.

Q: Construct the confidence interval for the population mean y. C=0.95, x= 16.7, o= 4.0, and n= 75. A…

A: Here, C=0.95, x=16.7, σ=4.0, and n=75.

Q: If n=19, (x-bar)=32, and s=11, construct a confidence interval at a 98% confidence level. Assume the…

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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 = 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: Find the margin of error for the 95% confidence interval used to estimate the population proportion.…

A: Confidence level = 95% Sample size (n) = 2440 Sample proportion (p̂) = 0.70

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: For a confidence level of 80% with a sample size of 24, find the t-score.

A: From the provided information, Confidence level = 80% Level of significance (α) = 1 – 0.80 = 0.20…

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 1850, 1970 is a confidence interval for a population mean. Find the value of the point estimator…

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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: Find the condence level and a for a. a 90% confidence interval.b. a 99% confidence interval.

A: a.The confidence level and α for a 90% confidence interval is obtained below:The confidence level is…

Q: sample of 35 students was taken from the local university. These students spent an average of $180…

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Q: A history teacher records the SAT scores of all 87 of her students. The data is used to construct a…

A: For calculating the confidence interval, we need to satisfy the following conditions: Independence…

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: Find 95% confidence interval for the population mean, if the sample size is 40, the sample mean is…

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Q: Construct the confidence interval for the population mean y. C= 0.90, x=16.6, o= 10.0, and n= 80

A: Here, C=0.90, x=16.6, σ=10.0, and n=80.

Q: Determine if you should use a confidence interval for the mean or a confidence interval for…

A: The aim is to determine whether to use confidence interval for mean or a confidence interval for the…

Q: own residents were interviewed and it was found that 422 responded favorably to a question while 501…

A: given data, 422-favorably responded501-negetively responded 923- total  interviewed for positive…

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: The number of successes and the sample size are given for a simple random sample from a population.…

A: Obtain the value of sample proportion. The value of sample proportion is obtained below: From the…

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 = 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: Construct the confidence interval for the population mean . C=0.98, x=15.3, = 10.0, and n=75

A: The Confidence Interval formula is X  ±  Z s √(n) Where: X is the mean Z is the Z-value…

Q: Which of the following confidence intervals would have a wider width for the same set of data? A 90%…

A: We have to find which of the following confidence intervals would have a wider width for the same…

Q: If n=30, 7(x-bar)=49, and s=14, construct a confidence interval at a 99% confidence level. Assume…

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Q: The County Development Agency does a survey of all the mobile home manufacturers in Clarion County…

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Q: FIND t a n = 15 for the 98% confidence interval for the mean.

A: Provided information that, n=15 Alpha (α)=1-0.98=0.02 α/2=0.01

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=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 X= 102, o=9, and n=63, construct a 95% confidence interval estimate of the population mean, μ.

A: Formula to constructing the confidence interval about mean , when the population standard deviation…

Q: Construct a 99​% confidence interval of the population proportion using the given information.…

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Q: If Xbar=93​, σ=13​, and n=62​, construct a 99​% confidence interval estimate of the population​…

A: Given that  Sample size: n=62 Sample mean: x¯=93 Population standard deviation: σ=13 Confidence…

Q: A researcher examined the extent of cooperation among kindergarten children. She unobtrusively…

A:

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 X=72​, σ=13​, and n=66​, construct a 95​% confidence interval estimate of the population​…

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Q: You measure 37 textbooks' weights, and find they have a mean weight of 73 ounces. Assume the…

A: From the provided information, Sample size (n) = 37 Sample mean (x̅) = 73 Population standard…

Q: In a survey of 2000 adults in a recent year, 713 made a New Year's resolution to eat healthier.…

A: We have given that  Sample size n= 2000 Favorable cases x = 713 Sample proportion p^=x/n

Q: Assume that a sample is used to estimate a population proportion p. To find the 95% confidence…

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Q: A local health clinic surveys 50 patients about their water drinking habits. It found the data is…

A: n=50 Mean=62 Standard deviation=5.2 Confidence level=95%

Q: A school district wants to know how many students in the district have peanut allergies. Use a 90%…

A: Standard deviation=σ=9 Margin of error=E=0.8

Q: If n=28, ¯x(x-bar)=32, and s=11, construct a confidence interval at a 95% confidence level. Assume…

A: