43. μ 1. (Confidence interval for µ given 02²) Determine the confidence interval [01, 02] μ,0.95 with confidence level y 0.95 for the mean μ of a normal distribution with variance σ² = 4, using a samples of n = 400 values with sample mean = 3. = 2. (Confidence interval for μ with unknown σ²) Six independent measurements are taken from a normal population. The sample has values x1 = 21, x2 = 19, x322, x4 = 20, x5 = 18, and x6 = 24. Find a 99% confidence interval for the mean.

A population has a mean u = 141 and a standard deviation o = 28. Find the mean and standard deviation of the sampling distribution of sample means with sample size n=56. The mean is H= and the standard deviation is o, =- (Round to three decimal places as needed.)

A population has a mean μ = 157 and a standard deviation o=20. Find the mean and standard deviation of the sampling distribution of sample means with sample size n = 42. The mean is μ =, and the standard deviation is o-= (Round to three decimal places as needed.)

The lengths of time​ (in years) it took a random sample of 32 former smokers to quit smoking permanently are listed. Assume the population standard deviation is 5.9 years. At α=0.03​, is there enough evidence to reject the claim that the mean time it takes smokers to quit smoking permanently is 15 ​years? Complete parts​ (a) through​ (e).   The lengths of time​ (in years) it took a random sample of 32 former smokers to quit smoking permanently are listed. Assume the population standard deviation is 5.9 years. At α=0.03​, is there enough evidence to reject the claim that the mean time it takes smokers to quit smoking permanently is 15 ​years? Complete parts​ (a) through​ (e).     19.9 21.3 13.1 17.8 17.5 18.3 7.3 18.1   18.9 9.5 17.5 19.4 15.3 16.5 7.6 18.7   22.8 18.6 21.3 11.5 12.7 18.3 13.5 20.7   12.7 12.1 14.3 16.9 19.9 21.2 22.9 10.3   ​(a) Identify the claim and state the null hypothesis…

Suppose we have the short hand info on a Normal Distribution as N(10,25). What is the population mean of the distribution and what is the population standard deviation? μ =  σ =

A population has mean  =μ19  and standard deviation  =σ5 . Find  μx  and  σx  for samples of size  =n25 . Round your answers to one decimal place if needed.     μx =     σx =

A random sample of 23 adult male wolves from the Canadian Northwest Territories gave an average weight x1 = 98.6 pounds with estimated sample standard deviation s, = 7.3 pounds. Another sample of 23 adult male wolves from Alaska gave an average weight x, = 88.2 pounds with estimated sample standard deviation s2 = 6.4 pounds. (a) Categorize the problem below according to parameter being estimated, proportion p, mean u, difference of means µ1 - 42, or difference of proportions p1 - P2. Then solve the problem. O P1 - P2 O H1 - 42 (b) Let µz represent the population mean weight of adult male wolves from the Northwest Territories, and let u2 represent the population mean weight of adult male wolves from Alaska. Find a 90% confidence interval for u1 - H2. (Use 1 decimal place.) lower limit upper limit (c) Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At…

What is tα/2,df for a 95% confidence interval of the population mean based on a sample of 15 observations?

Pressure values measured by various probes in the test section of a wind tunnel to obtain the freestream velocity have been given in the table below. The researcher who made the measurements concluded that the 7th reading had a value of -2.567 as the "deviation from the mean" and eliminated the data point based on the Chauvenet criterion. However, she later realized that she used the factor "1/n" instead of "1/(n-1)" in the calculation of standard deviation, which was used for data sets with a total number of data points less than 20. Find the value of the 7th reading. n=1 p=246 n=2 p=240 n=3 p=242 n=4 p=245 n=5 p=245 n=6 p=242 n=7 p=? n=8 p=244 n=9 p=244

n= 50 Mean, = ΣΧ = 1570/50 = 31.4 Standard deviation, o = = 20.41 Using empirical rule, 68% of the data falls within one standard deviation, 95% within two standard deviations, and 99.7% within three standard deviations from the mean. => 68% of data falls between e + o = 31.4 + 20.41 = (10.99, 51.81) => 95% of the data falls between e ± 2 * o = 31.4 + 2*20.41 = (-9.42, 72.22) => 99.7% of the data falls between + 3 * o = 31.4 +3*20.41 = (-29.83, 92.63) QUESTIONS: 1. How does answers in 68%, 95%, and 99.7% of data compare, respectively? 2. How does it support the conclusion you reached in question 1? Explain thoroughly.