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 112, and the sample standard deviation, s, is found to be 10. (a) Construct a 96% confidence interval about ? if the sample size, n, is 25. (b) Construct a 96% confidence interval about ? if the sample size, n, is 11. (c) Construct a 95% confidence interval about ? if the sample size, n, is 25. (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed?

Transcript

00:01 In this question, the sample mean given here, which is x bar, so this is 112, and the standard division for the sample, which was given as 10 here.

00:09 So in the first part of the question, the confidence level was given as 96 percent and the sample size, which is 25.

00:16 First of all, let me just write the formula for the confidence interval for the population mean.

00:23 So the confidence interval for the population mean, this is sample mean, plus or minus, because we know the sample standard division, use the t distribution, and sample standard division divided by root n.

00:33 Let me get the t value here.

00:34 So the alpha is 1 minus confidence level.

00:37 I need alpha over 2, that means this is 1 minus 0 .96, which is the 96 percent, divided by 2 would be 0 .02.

00:44 And also, i need the degrees of freedom, which is n minus 1 here, so this is 25 minus 1, which would be 24.

00:52 To get the t value here, what am i supposed to do? i'm going to use the inverse t function.

00:55 This is 0 .02, and the degrees of freedom, which is 24.

00:59 Press second, variance, and the inverse t here.

01:02 This is 0 .02, and then 24.

01:05 So the value would be negative 2 .172 here.

01:10 In the last step, i'm going to put everything to the formula.

01:12 So the sample mean, 112 plus or minus 2 .172, and times standard division divided by square root of the sample size, which is 25.

01:22 112, this is plus 2 .172, and times, this is 10 divided by square root of 25, which is 5.

01:30 So the upper boundary would be 116 points.

01:33 So how many decimal places we need? nothing given here.

01:36 Let me just write with one decimal places.

01:39 And the lower boundary, i'm going to get the same expression, just change the operation between them, which would be 107 and 0 .7 here.

01:47 This is the interval for the first part.

01:50 And the b here, the confidence level is again, this is 96 percent, but the sample size given here, which is 11.

01:58 So the alpha over 2 is again 0 .02, but the degrees of freedom, 11 minus 1, which is 10.

02:03 Let's get the t value here.

02:05 Again, i'm going to use the inverse t function, 0 .02, and then 10.

02:09 Press second, variance, and the inverse t here.

02:12 This is 0 .02, and then 10...

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