A sample of size n = 90 is drawn from a normal population...

A sample of size $n = 90$ is drawn from a normal population whose standard deviation is $sigma = 5.3$. The sample mean is $overline{x} = 37.79$. Part 1 of 2 (a) Construct a 98% confidence interval for $mu$. Round the answer to at least two decimal places. A 98% confidence interval for the mean is $Box < mu < Box$. Part 2 of 2 (b) If the population were not approximately normal, would the confidence interval constructed in part (a) be valid? Explain. The confidence interval constructed in part (a) (Choose one) be valid since the sample size (Choose one) large.

Transcript

00:01 We've been given some sample data.

00:02 Sample size n is 90.

00:05 Sample mean x bar is 37 .79.

00:09 We also have the population standard deviation, sigma, 5 .3.

00:15 Using this we want to make a 98 % confidence interval.

00:19 So the formula we need is point estimate x bar plus and minus the margin of error, z sigma over root n.

00:29 Sigma over root n is the standard error, z is our critical value.

00:33 There's another type of critical value, t, and why are we using z? well, z is preferable and we meet the requirements.

00:42 A big requirement is sigma should be known.

00:44 We do know sigma so we can definitely use z, but even if i didn't know sigma, even if i had s instead, i would still be using z here.

00:54 Because another possible requirement is having a large sample.

00:59 Different textbooks have different ideas what large means.

01:02 I use n over 30 because as sample size increases, s over root n becomes a better and better estimate of the standard error.

01:15 So what is z? we get it from a level of confidence and i'm going to explain this and explain part two as well while i'm at it.

01:25 As sample means, as sample size increases, the sample means get more and more normally distributed compared to the population.

01:34 That's the central limit theorem.

01:39 And if the sample size n is at least 30, you can treat the distribution of sample means as approximately normal regardless of the shape of the original distribution...

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