1. If n=18, ¯x(x-bar)=49, and s=4, find the margin of error at a 99% confidence level Give your answer to two decimal places. 2. If n=15, ¯x=49, and s=4, construct a confidence interval at a 90% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place. < μ <
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A: We have given that, Sample size (n) = 20, sample(x-bar) = 46, standard deviation (s) = 6 We wil…
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A: The given values are n =31, x̅ = 32, and s = 6.
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A: Since the population standard deviation is unknown, t distribution can be used for the computation…
Q: If n=32, x(x-bar)=47, and s=18, construct a confidence interval at a 99% confidence level. Assume…
A: Given that Sample size n =32 Sample mean =47 Standard deviation =18
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Q: If n=14, ¯xx¯(x-bar)=44, and s=9, construct a confidence interval at a 98% confidence level. Assume…
A: Given data ; n =14 , x̄ = 44 , Standard deviation (s) = 9 To construct 98 % confidence interval.
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A: The sample size is 28 The sample mean is 32 The sample standard deviation is 2.
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Q: If n=25, x(x-bar)=41, and s=9, construct a confidence interval at a 95% confidence level. Assume the…
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A: Given: n=60x=97σ=12Confidence Level =90%
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A: n = 29, x= 49, and s = 11 Confidence Interval : x±t0.98 sn = 49…
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A: Given problem Given that n = 12 x-bar = 35 s = 8 Construct a confidence interval at a 90%…
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A: Given data, n=32 x bar =37 s=8 α=0.10
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Q: If n = 15, π = 47, and s = 13, construct a confidence interval at a 95% confidence level. Assume the…
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Q: Pavan
A: Step 1: Step 2: Step 3: Step 4:
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A: n = 29x¯ = 35s = 12
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Q: If n = 16, ¯xx¯ = 45, and s = 11, construct a confidence interval at a 99% confidence level. Assume…
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Q: If n=31, ¯xx¯(x-bar)=45, and s=8, construct a confidence interval at a 99% confidence level. Assume…
A: Given: n=31x¯=45s=8
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A: given data n = 31 x¯ =38 s=16 normal distribution 98% ci for μ= ? α=1-0.98=0.02 Df = n-1 = 30
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Q: If n = 27, x = 48, and s= 14, construct a confidence interval at a 99% confidence level. Assume the…
A: Given n=27, sample mean x̄=48, s=14
Q: If n = 14, ¯xx¯ = 44, and s = 18, construct a confidence interval at a 95% confidence level. Assume…
A: Given information- Sample size (n) = 14 Mean x̅ = 44 Standard deviation (s) = 18 Significance…
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A: The degrees of freedom is, df=n-1=20-1=19 The degrees of freedom is 19. Computation of critical…
Q: If n=25, (x-bar)=47, and s=6, construct a confidence interval at a 95% confidence level. Assume the…
A: given data n = 25x¯ = 47s = 695 % ci for μ = ?
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Q: If n=30, (x-bar)=45, and s=13, construct a confidence interval at a 95% confidence level. Assume the…
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1. If n=18, ¯x(x-bar)=49, and s=4, find the margin of error at a 99% confidence level
Give your answer to two decimal places.
2. If n=15, ¯x=49, and s=4, construct a confidence interval at a 90% confidence level. Assume the data came from a
Give your answers to one decimal place.
< μ <
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- If n = 23, ¯x = 35, and s = 3, construct a confidence interval at a 95% confidence level. Assume the data came from a normally distributed population. Give your answers to three decimal places.< μ <If n=14, x(x-bar)=40, and s=10, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place.The mean diastolic blood pressure for a random sample of 50 people was 84 millimeters of mercury. If the standard deviation of individual blood pressure readings is known to be 9 millimeters of mercury, find a 90% confidence interval for the true mean diastolic blood pressure of all people. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.) Lower limit:| ? Upper limit:|If n=21, ¯x(x-bar)=31, and s=7, construct a confidence interval at a 99% confidence level. Assume the data came from a normally distributed population.Give your answers to one decimal place.< μ <If n=24, (x-bar)=44, and s=18, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place.A statistics class is estimating the mean height of all female students at their college. They collect a random sample of 42 female students and measure their heights. The mean of the sample is 65.3 inches and the sample standard deviation is 5.4 inches. Find the 90% confidence interval for the mean height of all female students in their school? Assume that the distribution of individual female heights at this school is approximately normal. 1. The t value is 1.3025 2. The standard error has a value of -- 3. The margin of error has a value of 4. Find the 90% confidence interval for the mean height of all female students in their schoolIf n=27, x¯(x-bar)=41, and s=3, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population.Give your answers to one decimal place. _________< μ< _______If n=27, (x-bar)=46, and s=19, construct a confidence interval at a 99% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place. <μ<If n=27, ¯xx¯(x-bar)=43, and s=14, construct a confidence interval at a 99% confidence level. Assume the data came from a normally distributed population.Give your answers to one decimal place. < μμ <If n=29, (x-bar)3D46, and s=7, construct a confidence interval at a 95% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place. < µ <If n = 13, x = 49, and s = 11, construct a confidence interval at a 99% confidence level. Assume the data came from a normally distributed population. Give your answers to three decimal places.If n=26, ¯xx¯(x-bar)=45, and s=6, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population.Give your answers to one decimal place. < μμ <SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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