Question

Use technology and the given confidence level and sample data to find the confidence interval for the population mean μ. Assume that the population does not exhibit a normal distribution. Weight lost on a diet: 99% confidence n=41 x overbar =4.0 kg s=6.9 kg What is the confidence interval for the population mean μ? ___ kg<μ<___ kg (Round to one decimal place as needed.) Is the confidence interval affected by the fact that the data appear to be from a population that is not normally distributed? A. No, because the population resembles a normal distribution. B. No, because the sample size is large enough. C. Yes, because the population does not exhibit a normal distribution. D. Yes, because the sample size is not large enough.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 4.0 kg

sample standard deviation = s = 6.9 kg

sample size = n = 41

Degrees of freedom = df = n - 1 = 41 - 1 = 40

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t_{0.005, 40} = 2.704

Margin of error = E = t_/2,df * (s /n)

= 2.704 * (6.9 / 41)

Margin of error = E = 2.9

The 99% confidence interval estimate of the population mean is,

- E < < + E

4.0 - 2.9 < < 4.0 + 2.9

( 1.1 kg < < 6.9 kg )

A. No, because the population resembles a normal distribution.