Construct a confidence interval for ? assuming that each...

Construct a confidence interval for ? assuming that each sample is from a normal population. (a) x? = 30, ? = 5, n = 15, 90 percentage confidence. (Round your answers to 2 decimal places.) The 90% confidence interval is from to (b) x? = 121, ? = 7, n = 25, 99 percentage confidence. (Round your answers to 2 decimal places.) The 99% confidence interval is from to (c) x? = 12.8, ? = 1.9, n = 45, 95 percentage confidence. (Round your answers to 3 decimal places.) The 95% confidence interval is from to

Transcript

00:01 We have to find the confidence interval for the population mean for each case here.

00:05 So for the part a, the sample mean is given here as 30, and the population standard deviation 5, the sample size is 15, and the confidence level which was given as 90%.

00:17 So let me write the formula, the confidence interval for the population mean, which is the sample mean, plus or minus, because we know the population standard deviation, we have to use the z distribution here, and the population standard deviation divided by root n.

00:30 Let me get the z value first, so the alpha is 1 minus confidence level, we need alpha over 2, this is 1 minus 0 .98, which is the 90%, divided by 2, that would be 0 .05.

00:41 To get the z value here, i'm going to use the graphing display calculator application, inverse norm, so the area, the mean, and the standard deviation for the standard normal distribution.

00:52 Press 2nd, variance, and the inverse norm here.

00:55 This is 0 .05, and then the mean and the standard deviation, so the value would be negative 1 .645.

01:03 In the next step, i'm going to put these values on the formula.

01:05 So the sample mean, 30 plus or minus 1 .645, and times the population standard deviation divided by the square root of the sample size, which is 15.

01:14 So this is 30 plus 1 .645, and times, this is 5, divided by the square root of 15, so the upper boundary would be with two decimal places, 32 .12.

01:28 Let me get the lower boundary.

01:29 For the lower boundary, i'm going to use the same expression, just use the minus sign here.

01:34 So this is 27 .88.

01:36 This is the interval for the part a.

01:39 What about for part b? in this case, the x bar is 121, 7, and is 25, and the confidence level, which is 99%.

01:48 Again, i need the alpha over 2, which is 1 minus 0 .99, divided by 2, would be 0 .005.

01:55 To get the z value here, because again we have the population standard deviation, so i'm going to use the inverse norm and the area, the mean and the standard deviation.

02:05 So press second variance and the inverse norm here.

02:08 This is 0 .005, and mean and standard deviation.

02:12 So the value would be negative 2 .576.

02:15 Let me put these values on the formula...

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