If n=24, x(x-bar)%=D30, and s=7, 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.
Q: If n=19, x¯(x-bar)=50, and s=2, find the margin of error at a 80% confidence level Give your answer…
A: x¯ = 50n = 9s = 280% confidence level
Q: If n=32, ¯xx¯(x-bar)=31, and s=14, construct a confidence interval at a 95% confidence level. Assume…
A: Given: x¯=31n=32s=1495% confidence level
Q: If n-29, 7(x-bar)-35, and s-7, construct a confidence interval at a 90% confidence level. Assume the…
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Q: If n=16, ¯xx¯(x-bar)=46, and s=12, construct a confidence interval at a 95% confidence level. Assume…
A: Given n=16 x=46 s=12 CL=95% Data is normally distributed
Q: If n=14, ¯xx¯(x-bar)=40, and s=4, find the margin of error at a 95% confidence level Give your…
A: Given that, Sample size = 14 Mean = 40 Standard deviation = 4 95% confidence level
Q: A If n=21, ¯x (x-bar)=34, and s=8, construct a confidence interval at a 98% confidence level. Assume…
A:
Q: If n=22, ¯xx¯(x-bar)=45, and s=4, construct a confidence interval at a 90% confidence level. Assume…
A: Ans - Given the level of confidence interval 90 % Then α = 1 - 0.90 =…
Q: If n = 22, x¯ = 34, and s = 19, construct a confidence interval at a 98% confidence level. Assume…
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Q: If n=31, ¯xx¯(x-bar)=30, and s=10, construct a confidence interval at a 80% confidence level. Assume…
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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 19, x =40, and s = 20, construct a confidence interval at a 80% confidence level.
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Q: If n=20, (x-bar)=D46, and s=6, find the margin of error at a 95% confidence level Give your answer…
A: We have given that, Sample size (n) = 20, sample(x-bar) = 46, standard deviation (s) = 6 We wil…
Q: If n = 22, ¯xx¯ = 44, and s = 11, construct a confidence interval at a 95% confidence level. Assume…
A: Obtain the 95% confidence interval for the population mean. The 95% confidence interval for the…
Q: If n=19, ¯xx¯(x-bar)=37, and s=13, construct a confidence interval at a 99% confidence level. Assume…
A:
Q: If n=11, ¯x(x-bar)=35, and s=8, find the margin of error at a 90% confidence level Give your answer…
A: Given that, n=11, x-bar=35, and s=8, Confidence level is 90%
Q: If n = 25, ¯x = 50, and s = 19, construct a confidence interval at a 99% confidence level. Assume…
A:
Q: If n=24, x(x-bar)=33, and s=3, find the margin of error at a 90% confidence level Give your answer…
A: It is given that standard deviation (s) is 3 and sample size (n) is 24.
Q: If n=12, =49, and s=8, construct a confidence interval at a 80% confidence level. Assume the data…
A:
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
Q: (c) How large would the sample need to be so that a 95% confidence interval would have a margin of…
A: Given that sample mean xbar=12.05 and population standard deviation=6.55 3) find sample size of…
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: Maria has data from a random sample of 16 subjects and is constructing a 95% confidence interval for…
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Q: If a confidence interval has a width 1 based on a sample of 50 observations, what is the width if…
A: Sampl size=50 , Width=1 If sample size is increased to 800 sigma, Zc are constant
Q: . A researcher collects a sample of 24 measurements from a population and wishes to find a 99%…
A: The degrees of freedom is,
Q: If n=22, ē(x-bar)=42, and s=19, construct a confidence interval at a 80% confidence level. Assume…
A: Solution: From the given information, n=22, x-bar=42 and s=19.
Q: If n=13, ¯x(x-bar)=41, and s=5, find the margin of error at a 80% confidence level Give your answer…
A:
Q: If n=n= 22, x= 32, and s= 16, construct a confidence interval at a 99% confidence level. Assume the…
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Q: If n=10, ¯x(x-bar)=38, and s=5, find the margin of error at a 80% confidence level Give your answer…
A:
Q: If n=20, (x-bar)=47, and s=15, construct a confidence interval at a 99% confidence level. Assume the…
A: The sample size is 20, sample mean is 47, and sample standard deviation is 15. The degrees of…
Q: If n = 12, ¯x = 47, and s = 19, construct a confidence interval at a 99% confidence level. Assume…
A:
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=72, σ=13, and n=66, construct a 95% confidence interval estimate of the population…
A:
Q: if n=29, x(x-bar)=45, and s=6 find the margin of error at 80% confidence level. give your answer to…
A: Obtain the margin of error at the 80% confidence level. The margin of error at the 80% confidence…
Q: If n=19, ¯xx¯(x-bar)=38, and s=8, construct a confidence interval at a 90% confidence level. Assume…
A: Given, Sample size = 19 Sample mean = 38 Sample standard deviation = 8 Since the population…
Q: If n=30, (x-bar)=45, and s=13, construct a confidence interval at a 95% confidence level. Assume the…
A: We have given that, Sample mean (x̄) = 45, sample size (n) = 30 and standard deviation (s) = 13…
Q: If n = 24 , overline x (x-bar)=40 , and s = 20 . construct a confidence interval at a 95% confidence…
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Q: If n=31, ¯xx¯(x-bar)=33, and s=9, construct a confidence interval at a 80% confidence level. Assume…
A:
Q: If n=13, (x-bar)=43, and s=7, construct a confidence interval at a 99% confidence level. Assume the…
A: The provided sample mean is X¯=43 and the sample standard deviation is s=7. The size of the sample…
Q: Given that you need to construct a 90% confidence interval, having 12 measurements, and you don't…
A: The confidence level is c = 0.90.
Q: If n=16, ¯xx¯(x-bar)=45, and s=17, construct a confidence interval at a 99% confidence level. Assume…
A:
Q: If X=68, σ=14, and n=66, construct a 95% confidence interval estimate of the population mean, μ
A:
Q: If n-20, 7(x-bar)=31, and s=17, construct a confidence interval at a 95% confidence level. Assume…
A: Given that n =20 ,s =17 ,x bar =31 ,alpha =0.05
Q: If n=16, 7(x-bar)=38, and s=12, construct a confidence interval at a 99% confidence level. Assume…
A: Since population standard deviation is unknown, Use t-distribution to find t-critical value. Find…
Q: If n=20, ¯xx¯(x-bar)=43, and s=10, construct a confidence interval at a 98% confidence level. Assume…
A: Here we need to compute a 98% confidence interval from the given information. Given sample size=20…
Q: If n=13, x(x-bar)=31, and s=4, find the margin of error at a 95% confidence level Give your…
A: n = 13 Mean = 31 Sample standard deviation = 4 95% confidence level
Q: If n=16, x¯(x-bar)=48, and s=7, construct a confidence interval at a 98% confidence level. Assume…
A:
Q: If n=22, 7(x-bar)=34, and s=17, construct a confidence interval at a 99% confidence level. Assume…
A: According to the given information, we have Sample size, n = 22 Sample mean = 34 Sample standard…
Q: If n=16, ¯xx¯(x-bar)=45, and s=7, find the margin of error at a 95% confidence level Give your…
A:
Q: If n=24, (x-bar)=D34, and s=9, construct a confidence interval at a 95% confidence level. Assume the…
A:
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- An SRS of 350 high school seniors gained an average of ?⎯⎯⎯=45points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation ?=26.deviation Find a 90% confidence interval for ?� based on this sample. Give your answers to three decimal places. What is the lower bound? What is the upper bound ?A sample of size n = 80 is drawn from a normal population whose standard deviation is σ = 6.8. The sample mean is x̄ = 40.41. a. Construct a 90% confidence interval for μ. b. If the population were not approximately normal, would the confidence interval constructed in part (a) be valid? Explain.A sample of size n=70 is drawn from a normal population whose standard deviation is σ=5.1 . The sample mean is x=35.46 (a) Construct a 98% confidence interval for μ. Round the answer to at least two decimal places. (b) If the population were not approximately normal, would the confidence interval constructed in part (a) be valid?Explain.
- A sample of size n = 85 has sample mean of 81 and sample standard deviation s = 5.9. i. Construct a 95% confidence interval for the population mean μ. ii. If the confidence level were 99%, would the confidence interval be narrower or wider?A random sample of 10 items is taken from a normal population. The sample had a mean of 82 and astandard deviation is 26. Which is the appropriate 99% confidence interval for the population mean?A random sample of 250 preschool children in a local suburb revealed that only 232 had been vaccinated. Provide an approximate 95% confidence interval for the proportion vaccinated in that suburb. You MUST show your work to receive full credit. Give an interpretation of the confidence interval.
- A random sample of 100 observation produced a sample mean 22.3 and given population standard deviation 6. (b) find a 90% confidence interval for μ. Round to 2 decimal places.A sample of 31 cans of soup have a sample average of 102.1 mg of sodium with a sample standard deviation of 4.7 mg of sodium. Construct a 95% confidence interval for the mean sodium for all Seinfield soup. Also find the margin of error. Round your answers to 3 decimal places.Show work and curves Round to 2 decimals If the mean annual salary for 20 randomly selected nurses is $39000, find the 90% confidence interval for the population mean salary if sigma is $1700.
- How large a sample should be selected to provide 95% confidence interval with a margin of error 8? Assume that the population standard deviation is 50. Round your answer to next whole number.NEED ASAP. A pharmaceutical claimed that a new drug introduced in the market could help women reduce weight by 7 kilograms per month with standard deviation of 1 kilogram. A sample of 25 women were randomly chosen and found out that their weight at an average 6.20 kilogram in a span of one month. Does this data make the claim of the manufacturer valid at 95% confidence level?Breaking strengths of a random sample of 28 molded plastic housing were measured, and a sample standard deviation of S=19.52 was obtained. A confidence interval (439.23, 460.77) for the average strength of molded plastic housings were constructed from these results. What is the confidence level of this confidence interval?