I want answers with explaination. A random sample of n = 49 observations has a m

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I want answers with explaination.

A random sample of n = 49 observations has a mean x = 28.6 and a standard deviation s = 3.4. (a) Give the point estimate of the population mean, mu. Find the 95% margin of error for your estimate. (Round your answer to four decimal places.) (b) Find a 90% confidence interval for mu. (Round your answers to three decimal places.) What does "90% confident" mean? In repeated sampling, 90% of all intervals constructed in this manner will enclose the population mean. There is a 10% chance that an individual sample mean will fall within the interval limits. There is a 90% chance that an individual sample mean will fall within the interval. 90% of all values will fall within the interval limits. In repeated sampling, 10% of all intervals constructed in this manner will enclose the population mean. (c) Find a 90% lower confidence bound for the population mean mu Why is this bound different from the lower confidence limit in part (b)? This bound is calculated using z_alpha/2, while the lower confidence limit in part (b) is calculated using z_alpha. This bound is calculated using n, while the lower confidence limit in part (b) is calculated using squareroot n. This bound is calculated using squareroot n while the lower confidence limit in part (b) is calculated using n. This bound is calculated using z_alpha while the lower confidence limit in part (b) is calculated using z_alpha/2. The lower bounds are based on different values of x. (d) How many observations do you need to estimate mu to within 0.4, with probability equal to 0.95? (Round your answer up to the nearest whole number.) observations

Explanation / Answer

a) Point estimate of the population mean x = 28.6

z value = 1.96

ME = z * s / sqrt(n)

= 1.96 * 3.4 / sqrt(49)

= 0.952

b) z value = 1.645

CI = mean + / - z * ( s/sqrt(n))

= 28.6 + /- 1.645 * ( 3.4/sqrt(49))

= (27.801 , 29.399)

This CI indiacates that there is 90% chance that the individual sample mean will fall within the interval.

c) In order to determine the lower bound we use z = -1.28. This is one sided confidence interval.

lower bound = mean + z*sigma/sqrt(n)

lower bound = 28.6 - 1.28*3.4/sqrt(49)

lower bound = 27.9783

This is because this bound is calculated using z(alpha), while the lower confidence limit in part (b) is calculated using z(alpha/2)

(d)

ME = z*SE

0.4 = 1.96*3.4/sqrt(n)

n = (1.96*3.4/0.4)^2 = 277.56

Hence we need 278 observations

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