A simple random sample of size n is drawn from a population...

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found to be 111, and the sample standard deviation, s, is found to be 10. (a) Construct a 96% confidence interval about mu if the sample size, n, is 23. (b) Construct a 96% confidence interval about mu if the sample size, n, is 27. (c) Construct a 98% confidence interval about mu if the sample size, n, is 23. (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed?

Transcript

00:01 Let me take note the given values here.

00:03 So first of all, the sample mean, which is common for both cases, this is 111, and the sample standard division, that was given as 10 here.

00:12 And in the first part of the question, the confidence level, which was given as 96%, so we can write as 0 .96, and the sample size given as 23.

00:22 So we have to find the confidence interval for this case.

00:24 Let's remember the formula.

00:26 The confidence interval for the mean, because we know the sample standard division, we have to use the t distribution here.

00:32 So this is the sample mean, plus or minus the t distribution times sample standard division divided by root n.

00:38 So the alpha is 1 minus confidence level, but we need alpha over 2, 1 minus 0 .96 divided by 2, which would be 0 .02.

00:47 And the degrees of freedom, this is n minus 1 here, so this is 23 minus 1, which is 22.

00:53 Let's get the t alpha over 2 value, which is the inverse t function.

00:57 This is 0 .02, and degrees of freedom 22.

01:01 Press second variance and the inverse t here, this is 0 .02 and then 22.

01:06 So the value would be negative 2 .183.

01:09 Let me put these values on the formula.

01:11 So the sample mean, 111, this is plus or minus 2 .183, n times the sample standard division divided by root, which is 23.

01:22 111 plus 2 .183, n times, this is 10 divided by the square root of 23.

01:30 So the upper boundary would be 115 .55.

01:34 Let me get the lower boundary.

01:35 For the lower boundary, i'm going to use the minus sign here.

01:38 When i use the minus sign, this is 106 .45.

01:42 This is the interval we have.

01:45 And for part b.

01:46 So in this case, the confidence level is again 96%, but the sample size that was given as 27.

01:53 So the alpha over 2, which is 0 .02, degrees of freedom 27 minus 1, 26.

01:58 Let's get the t value here...

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