C. FIND t a n = 15 for the 98% confidence interval for the mean.
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: d. ta and n = 8 for the 95% confidence interval for the mean. 2 е. ta and n= 21 for the 85%…
A:
Q: If n=400 and X=100, construct a 95% confidence interval estimate for the population proportion.
A: Given,x=100n=400sample…
Q: d. ta and n = 8 for the 95% confidence interval for the mean. e. te and n= 21 for the 85% confidence…
A: Given that d) Confidence level = 95% = 0.95 Sample size (n) = 8 We know that Degrees of freedom…
Q: If X = 114, α= 20, and n = 33, construct a 99% confidence interval estimate for the population mean,…
A: Given data: Sample Mean = 114 Population standard deviation = 20 Sample size = 33 Confidence level…
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A: Instruction : "I need help with question 6 plz." Given Data :
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Q: Find a 95% confidence interval for a mean, µ, where n = 324, x̄ = 2.7, s = .91, and se = .06.
A: Given: n = 324, x̄ = 2.7, s = .91, se = .06.
Q: find the margin of error for a sample mean, given n=16, s=4, Xbar=17. for a 95% confidence interval.
A: The margin of error for sample mean is given by,
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A: We know that the Margin of error = (upper_limit - lower_imit )/2 and Sample mean = (upper limit +…
Q: Construct the indicated confidence interval for the population mean p using the t-distrib c=0.90, x=…
A: Solution: Given information: n= 9 Sample size x= 13.3 Sample mean s=2.0 Sample standard…
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A:
Q: Which of the below Z scores is the appropriate z* when calculating a 98% confidence interval? (a) Z…
A: We hvae to find corrcet answer..
Q: Construct the indicated confidence interval for the population mean . Assume c = 0.98, = 101.3, =…
A:
Q: For a two-tailed test, a sample size of 21 at 80% confidence, t= 1.323 2.539 1.325 2.528
A: We have to find correct option..
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: 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: Use the given degree of confidence and sample data to construct a confidence interval for the…
A: Given Information n =87 x =48
Q: ccording to the data, calculate a 90% confidence interval for the sample mean. What was the T Score…
A: Solution: n= 30 Sample size ∑xi=642 x=∑xin=64230=21.4 Sample mean s2=∑(xi-x)2n-1=56.52414 Sample…
Q: Use the given degree of confidence and sample data to construct a confidence interval for the…
A: According to the provided information: The proportion can be calculated as:
Q: Use the given confidence interval to find the margin of error and the sample proportion.…
A: According to the given information, we have Confidence interval = (0.690, 0.718)
Q: kendall wants to estimate the percentage of vegetarians who are also vegan. She surveys 150…
A: The margin of error for the 95% confidence interval of population proportion is given as follows :…
Q: Find the confidence level and α for a. an 85% confidence interval. b. a 95% confidence interval.
A:
Q: Assume the given confidence interval for the population mean height of Sequoia trees in Sequoia…
A: Given: Lower limit = 56.1 Upper limit =63.4 The Sample mean can be calculated as: Sample mean =…
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: f n=200 and X=80, construct a 99% confidence interval estimate for the population proportion.
A:
Q: d. Construct a 95% confidence interval estimate. How does this change your answer to part (b)?
A: Given that Sample size n = 50 Sample mean = 0.987 Population SD = 0.02 The critical value of Z at…
Q: If X=72, σ=13, and n=66, construct a 95% confidence interval estimate of the population…
A:
Q: Use the given degree of confidence and sample data to construct a confidence interval for the…
A: It is given that n=195 and x=162. The sample proportion is 162/195 = 0.8308 The sample size is 195…
Q: (b) Use a 99% confidence interval to estimate the mean percent change in the population. FILE O…
A: (b) The mean percent is, x¯=∑xn=-4.7-2.5-4.9-2.7+⋯-5.147=-168.647=-3.5872 The mean percent is…
Q: Use the Student's t distribution to find te for a 0.95 confidence level when the sample is 29. Step…
A: Given that the sample size is 29 and the confidence level is 0.95. That is, c=0.95 and n=29…
Q: Find the level of a two-sided confidence interval for t = 2.776 with sample size 5. Express the…
A: According to the provided information, sample size (n) = 5 The degrees of freedom is: df =n-1 = 5-1…
Q: find the t/α2 value for a 95% confidence interval for the mean when the sample size is 10
A:
Q: Find the confidence level and a for a. an 85% confidence interval.b. a 95% confidence interval.
A: The level of significance can be defined as the probability of rejecting the null hypothesis when it…
Q: Use the given confidence interval to find the margin of error and the sample mean. (4.81, 8.09)
A: The confidence interval is given by (4.81, 8.09)
Q: Find the confidence level and a for an 88% confidence interval.
A: Here use confidence level and significance level as alternative of each other
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A: We have to find margin of error
Q: In a poll conducted, 439 of 1007 adults in a certain country polled said they were "picky eaters."…
A: a. The proportion of the respondents said they were picky eaters is, p^=xn=4391007=0.4359≈0.44 Thus,…
Q: is: n=217, x 26.5 hg, s= 6.1 hg. Construct a confidence interval estimate of the mean. Use a 99%…
A: Given that Sample size(n) = 217 sample mean (x¯) = 26.5 Sample standard deviation (s) = 6.1…
Q: Construct a confidence interval for the population proportion p. n ? 144, x ? 86, 90% confidence
A: Given: Confidence level = 0.90 Sample size (n) = 144 x = 86 Sample proportion p^=xn=86144
Q: 300 high school students were asked how many hours.of TV they watch per day. The mean was 2 hours,…
A: Given that: Mean=2 Standard deviation=SD=σ= 0.5 Confidence level= 90% n=sample size=n=300 So…
Q: If X=68, σ=14, and n=66, construct a 95% confidence interval estimate of the population mean, μ
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Q: If [1740, 1850] is a confidence interval for a population mean. Find the margin of error. Select…
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Q: Find the confidence level and α for an 80% confidence interval.
A: Given confidence level =80%
Q: Use the given degree of confidence and sample data to construct a confidence interval for the…
A: Given data, n=195 x=126 p=x/n=126/195=0.646 z-value at 95% confidence is Zc=1.96
Q: (d) Use the combined data of 80 students, construct a 95% confidence interval estimate of the…
A: Given that Total students=n=40+40=80 Number of pass students=X=28+32=60
Q: find the value: t a 2 and n = 15 for the 98% confidence interval for the mean. t a 2 and n= 20…
A: a) Degrees of freedom=n-1=15-1=14 Then, at 98% confidence level, the value of t is 2.624. (From…
Q: c) i: Calculate the average and median to the results. ii: Find a 95% confidence interval for µ.
A: S.no. x (x-xbar)^2 1 11.79 0.03092 2 11.82 0.02127 3 11.87 0.00918 4 11.92 0.00210 5 11.93…
Q: Find the value of ta/2 given a. 37 for the 85% confidence interval for the mean b. n = 15 for the…
A:
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