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, is found to be 109, and the sample standard deviation, s, is found to be 10. (a) Construct a 95% confidence interval about μ if the sample size, n, is 22. (b) Construct a 95% confidence interval about μ if the sample size, n, is 18. (c) Construct a 90% confidence interval about μ if the sample size, n, is 22. (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 notes what values are given here.

00:03 So the sample mean which is denoted by x bar, this is 109, and the sample standard division denoted by s here, that was given as 10.

00:11 So for part a we're going to use the confidence level as 90 percent, so we can write that 0 .90, and the sample size which is given as 22 here.

00:20 So we're going to find the confidence interval for the population mean.

00:24 Let's remember the formula.

00:25 The confidence interval for the population mean, because we know the sample standard division, so we are going to use the t distribution.

00:33 Sample mean plus or minus t alpha over 2, sample standard division divided by square root of n.

00:39 Let's get the alpha value first.

00:40 This is 1 minus confidence level, but we need alpha over 2, 1 minus 0 .90, and divide by 2 which is 0 .05 here, and the degrees of freedom which is n minus 1 here, so this is 10 minus 1 which is 9.

00:52 In order to get the t alpha over 2, i'm going to use the inverse t function here.

00:57 So the area was 0 .05 and the degrees of freedom is 9 here.

01:01 Sorry, this is, so the, sorry, the degrees of freedom is 22 minus 1 which is 21.

01:09 My bad, this is 21 here.

01:11 Press second variance and the inverse t here, this is 0 .05 and the degrees of freedom is 21 which is negative 1 .72, and the last step i'm going to put all these findings to the formula.

01:23 So the sample mean plus or minus 1 .72, sample standard division divided by square root of this value.

01:31 This is 109 plus 1 .72 and times 10 divided by square root of 22.

01:39 So the upper boundary for this question which is 112 .67, and to get the lower boundary, let me get the same expression, just change the operation between them which is minus in this case.

01:51 So this is 105, 105 .33.

01:55 This is the answer for part a.

01:57 What about for part b? in this case, the confidence level is 95 percent, so i can write 0 .95, and the sample size given here which is 18.

02:07 So the alpha over 2, 1 minus 0 .95, divide by 2 which is 0 .025, and the degree of, degrees of freedom 18 minus 1 which is 17.

02:18 Let's get the t alpha over 2 which is inverse t.

02:22 This is 0 .025 and 17.

02:25 Press second variance and the inverse t here, 0 .025, and the degrees of freedom is 17.

02:32 So the value here negative 2 .11.

02:34 So i'm going to put this value again for the formula.

02:37 This is 109 plus or minus 2 .11 times, this is 10 divided by square root of, so the sample size here which is 18, i'm going to put the 18 here.

02:49 Let me get this expression here.

02:51 I'm going to change the t value to 0 .11 and the sample size which is 18 here.

02:58 So the lower boundary, this is 104 .03, and let's get the upper boundary...

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