A random sample of n 49 observations has a mean bar x 291

A random sample of n = 49 observations has a mean bar x = 29.1 and a standard deviation s = 3.1. (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.) to What does \'\'90% confident\'\' mean? 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. There is a 90% chance that an individual sample mean will fall within the interval. 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. (c) How many observations do you need to estimate mu to within 0.5, with probability equal to 0.95? (Round your answer up to the nearest whole number.) observations If it is assumed that the heights of men are normally distributed with a standard deviation of 2.5 Inches, how large a sample should be taken to be fairly sure (probability 0.95) that the sample mean does not differ from the true mean (population mean) by more than 0.70 in absolute value? (Round your answer up to the nearest whole number.) heights

Solution

(7)

(a) mu= 29.1

Given a=0.05, Z(0.025) =1.96 (from standard normal table)

So the margin of error is

Z*s/vn= 1.96*3.1/sqrt(49)=0.8680

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(b) Given a=0.1, Z(0.05) = 1.645 (from standard normal table)

So the lower bound is

xbar - Z*s/vn =29.1 -1.645*3.1/sqrt(49) =28.372

So the upper bound is

xbar +Z*s/vn = 29.1 +1.645*3.1/sqrt(49) =29.829

In repeated sampling, 90% of al intervals constructed in this manner will enclose the population mean

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(c)

n=(Z*s/E)^2

=(1.96*3.1/0.5)^2

=147.6711

Take n=148

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(8)

n=(Z*s/E)^2

=(1.96*2.5/0.7)^2

=49