A random sample of 93 observations produced a mean x = 25.5 and a standard deviation s 2.8. a. Find a 95% confidence interval for u. b. Find a 90% confidence interval for u. c. Find a 99% confidence interval for u.
Q: Assume that a sample is used to estimate a population mean μ. Find the 95% confidence interval for a…
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Q: Given the following measurements from a normal distribution: 8.6, 16.7, 18.5, 11.7, 9.7 a) Use the…
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Q: Assume that a sample is used to estimate a population mean μ. Find the 90% confidence interval for a…
A: Sample mean =44.1 Sample standard deviation, s = 7.9 The sample size, n is 45.
Q: Assume that a sample is used to estimate a population mean µ. Find the 99.5% confidence interval for…
A: Solution To calculate the confidence interval we will use t distribution.
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A: Given,sample size(n)=11standard deviation(s)=4.934α=1-0.95=0.05α2=0.0251-α2=1-0.025=0.975degrees of…
Q: Assume that a sample is used to estimate a population mean μμ. Find the 90% confidence interval for…
A: Given: Sample size (n) = 38 Sample mean x¯=39.2 Sample standard deviation (s) = 13.9 Confidence…
Q: 2. Construct a 95% confidence interval for the population standard deviation of a random sample of…
A: given data sample size(n) = 15 sample standard deviation(s) = 12.7construct 95 % ci for σ.
Q: Assume that a sample is used to estimate a population mean u. Find the 99.5% confidence interval for…
A: Confidence interval give a range of values for an unknown parameter of the population. The width of…
Q: Construct a 99% confidence interval for the population mean,μ.Assume the population has a normal…
A: Given information Sample Standard deviation = 3.8 Sample mean = 22.4 Sample size n = 19 Population…
Q: Assume that a sample is used to estimate a population mean μμ. Find the 99.9% confidence interval…
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Q: Assume that a sample is used to estimate a population mean μ. Find the 95% confidence interval for a…
A: It is given that Sample mean = 73.4 Sample standard deviation = 5.2 Sample size = 47 Confidence…
Q: Assume that a sample is used to estimate a population mean μμ. Find the 99.5% confidence interval…
A: given data sample size (n) = 36sample mean ( x¯ ) = 76.2sample standard deviation (s) =6.999.5% ci…
Q: Assume that a sample is used to estimate a population mean �. Find the 95% confidence interval for a…
A: It is required to find the 95% confidence interval to estimate the population mean.
Q: irate student complained. The cost of textbooks was too high. He randomly surveyed 36 other students…
A: As the population standard deviation is not known,. we will use z distribution. The value of z at a…
Q: Assume that a sample is used to estimate a population mean μμ. Find the 95% confidence interval for…
A: Given information Sample size (n) = 75 Sample mean = 83.2 Sample standard deviation = 11.7…
Q: Assume that a sample is used to estimate a population mean μμ. Find the 99.5% confidence interval…
A: IntroductionThe 100 (1 – α) % confidence interval for the population mean, μ, for given sample…
Q: The standard deviation of the diameter of 18 baseballs was found to be o.29 cm. Find the 95%…
A: Given: Sample size (n) = 18 Sample standard deviation (s) = 0.29 Confidence level = 0.95 So,…
Q: Assume that a sample is used to estimate a population mean u. Find the 90% confidence interval for a…
A: The sample size is 56, the sample mean is 19.4 and the sample standard deviation is 17.9.
Q: Assume that a sample is used to estimate a population mean u. Find the 80% confidence interval for a…
A: Givensample size(n)=52Mean(x)=55.2standard deviation(s)=9.1confidence inerval = 80%
Q: Assume that a sample is used to estimate a population mean μμ. Find the 95% confidence interval for…
A: Here we need to construct a 95% confidence interval for the population mean. Since the population…
Q: Assume that a sample is used to estimate a population mean u. Find the 99.5% confidence interval for…
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Q: Assume that a sample is used to estimate a population mean μμ. Find the 90% confidence interval for…
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Q: From a sample of 35 men, it was found that the mean weight was 175.2 lbs. and a population standard…
A: Use EXCEL Procedure for finding the critical value of z. Follow the instruction to obtain the…
Q: The mean and standard deviation of a random sample of n measurements are equal to 34.9 and 3.1,…
A: Mean = 34.9 stdev = 3.1
Q: Assume that a sample is used to estimate a population mean. Find the 90% confidence interval for a…
A: The summary of statistics is, x¯=42.9,s=20.4,n=62 The degree of freedom is, df=n-1 =62-1 =61…
Q: An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other…
A: Given that,x¯=121.60,σ=6.36,n=36 Critical value: By using the z-tables, the critical value at 10%…
Q: me that a sample is used to estimate a population mean μμ. Find the 90% confidence interval for a…
A: As the population standard deviation is not known, we will use t-distribution. Given: Xˉ=16.8 s=13.1…
Q: 7. Construct a 95% confidence interval for the population mean, μ. Assume the population has a…
A: CL=0.95 Sample size, n=20 Sample mean, x¯=3120 Sample sd, s=677
Q: Assume that a sample is used to estimate a population mean μμ. Find the 90% confidence interval for…
A: From the provided information, Sample size (n) = 53 Sample mean (x̅) = 56.8 Sample standard…
Q: Assume that a sample is used to estimate a population mean . Find the 95% confidence interval for a…
A: given data n = 41x¯ = 54.6σ = 7.395% ci for μ = ?
Q: Assume that a sample is used to estimate a population mean u. Find the 98% confidence interval for a…
A: Mean = 23.2 Standard deviation = 15.1 Sample size = 50
Q: The average height of students in a state from an SRS of 17 students gave a standard deviation of…
A: given data sample size(n) = 17 sample standard deviation(s) = 2.9construct 95 % ci for σ.
Q: The mean and standard deviation of a random sample of n measurements are equal to 34.4 and 3.6,…
A: Confidence interval: Confidence interval is a range of values it contains the true population…
Q: Assume that a sample is used to estimate a population mean μμ. Find the 99.5% confidence interval…
A: Solution: Let X be the random variable. From the given information, x-bar=41.7, S=12.3 and n=438.
Q: Assume that a sample is used to estimate a population mean μμ. Find the 98% confidence interval for…
A: Assume the population is normally distributed.
Q: Assume that a sample is used to estimate a population mean μμ. Find the 90% confidence interval for…
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Q: The sample mean for the fill weights of 100 boxes is 12.050. The population variance of the fill…
A: We want to find 95% confidence interval Sample Mean \bar XXˉ = 12.05012.050 Population…
Q: Assume that a sample is used to estimate a population mean μμ. Find the 90% confidence interval for…
A: We have given that Sample size n =39 Sample mean =74.7 Standard deviation =19.7
Q: Assume that a sample is used to estimate a population mean μμ. Find the 98% confidence interval for…
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Q: The mean weight of trucks traveling on a particular section of l-475 is not known. A state highway…
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Q: Assume that a sample is used to estimate a population mean μμ. Find the 98% confidence interval for…
A: Given Information: n=36Mean, x¯=58.3Standard Deviation, s=16.7Confidence Interval, c=98% =0.98
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- Over a period of 10weeks, a veterinarian took blood glucose reading from a horse, finding a mean of 93.8 mg/100 ml. and a standards deviation of 12.5 mg/ml. The mean glucose level for a healthy horse is 85 mg/ml. a. Perform a test of hypothesis at a 80% confidence level. Use the p-value method and a two tailed test of hypothesis. b. Perform a test of hypothesis at a 90% confidence level. Use the critical value method and a right tailed test.A sample of 16 ten-year-old girls had a mean weight of 71.5 and a standard deviation of 12 pounds, respectively. Assuming normality, find the 90 percent confidence intervals for mean the lower limit is ...Assume that a sample is used to estimate a population mean μ. Find the 95% confidence interval for a sample of size 44 with a mean of 49.5 and a population standard deviation of 18.5. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. 95% C.I. = The answer should be obtained without any preliminary rounding.
- The standard deviation, S, for the annual medical cost of 24 households was found to equal $1689 find a 95% confidence interval for the standard deviation of the annual medical cost for all households. Assume medical costs are normally distributed.Assume that a sample is used to estimate a population mean μ. Find the 90% confidence interval for a sample of size 69 with a mean of 82.5 and a standard deviation of 21.4. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. 90% C.I. = The answer should be obtained without any preliminary rounding.Assume that a sample is used to estimate a population mean μμ. Find the 80% confidence interval for a sample of size 55 with a mean of 85.5 and a standard deviation of 21.5. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places.a. 80% C.I. = The answer should be obtained without any preliminary rounding
- Twelve samples of coal from a Northern Appalachian source had an average mercury content of 0.253 ppm with a standard deviation of 0.029 ppm. Suppose the density of mercury content of coal from this source is approximately normally distributed. a. Find the LOWER bound of a 95% confidence interval for the mean mercury content of coal from this source. b. Find the UPPER bound of a 95% confidence for the mean mercury content of coal from this source.4. Find the 90% confidence interval for the variance and standard deviation for the time it takes a telephone company to transfer a call to the correct office. A sample of 15 calls has a standard deviation of 1.6 minutes. Assume the variable is normally distributed. 5.00Assume that a sample is used to estimate a population mean μμ. Find the 80% confidence interval for a sample of size 46 with a mean of 15.6 and a standard deviation of 10.2. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places.80% C.I. = The answer should be obtained without any preliminary rounding.
- Assume that a sample is used to estimate a population mean u. Find the 99% confidence interval for a sample of size 42 with a mean of 68,6 and a standard deviation of 21.9. Enter your answer as an open- interval (i.e., parentheses) accurate to 3 decimal places. 99% C.I. = The answer should be obtained without any preliminary rounding.You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.From a random sample of 66 dates, the mean record high daily temperature in a certain city has a mean of 85.22 F. Assume the population standard deviation is 15.45 F.A cellular battery manufacturer claims that his battery when fully charged has mean life of 24 hours with standard deviation of 4 hours. A dealer randomly chose sample of 35 batteries to be tested and resulted to 22.5 hours mean life. In the given situation, 22.5 hours is a. sample mean b. number of sample c. sample standard deviation d. population standard deviation