Problem 18. A sample of size 20 is drawn from a normal...

Problem 18. A sample of size 20 is drawn from a normal distribution with unknown variance and mean. The sample variance is $s^2 = 0.012$. Find a 95% two-sided confidence interval for the standard deviation $\sigma$ of the population. A. $[0.0833, 0.1600]$, B. $[0.010, 0.0130]$, C. $[0.0069, 0.0256]$, D. None of the above Problem 19. A sample data is drawn from a normal population with unknown mean $\mu$ and known variance. The $P$-value for the test $H_0: \mu = 10$ vs $H_1: \mu > 10$ is 0.05. Which of the following is correct. A. One would reject $H_0: \mu = 10$ in favor of $H_1: \mu > 10$ at the level 4% of significance B. One would reject $H_0: \mu = 10$ in favor of $H_1: \mu > 10$ at the level 5% of significance C. One would reject $H_0: \mu = 11$ in favor of $H_1: \mu > 10$ at the level 5% of significance D. None of the above Problem 20. The U.S. Postal Service has used optical character recognition. In a preliminary sample of 1000 handwritten zip code digits, 900 were read correctly. How large a sample is required if we want to be 95% confident that the error in using the sample proportion $\hat{p}$ to estimate the true proportion $p$ of correctly read zip code digits is less than 0.02? A. 864 B. 9 C. 100 D. 613

Transcript

00:01 You have lots of questions here on.

00:01 Your first question number four, you want to find what for a 96 .6 % confidence interval, that's going to give you a 0 .17 here in 0 .17, what the appropriate critical value is, and that is letter a, 2 .12.

00:22 For number five, you wanted to find if you had had a standard deviation of 10, and the standard deviation had been actually 20 instead.

00:38 So if that was the first time, what's going to happen to that confidence interval? and that confidence interval will be twice as wide.

00:45 So it will be 50 .92 plus or minus twice that value, which will be that 4 .28, because that standard.

00:54 Standard deviation over the square root of n times that particular z value or t value if it's a sample.

01:02 This is twice as big.

01:04 It's going to make that interval be twice as wide.

01:08 For number six, you have a sample size of 50, and if you increase that sample size to 200, what's going to happen to that interval? and again, that interval is going to actually end up.

01:20 It is four times bigger sample size and then square root that by 2.

01:25 Two, the margin of air will actually be half the size.

01:30 So the interval that is correct for that is letter b.

01:34 It will be that 19 .76 plus or minus, and then that margin of air will get cut in half.

01:41 Number seven, you want to know what sample size if you want the margin of air to be plus or minus two, and you have 1 .96.

01:50 It's a 95 % confidence interval, and you know that the standard deviation is eight.

01:55 And when you do the solving, this will round up to 62.

02:02 Number eight, you want to find what that mean is of the population, and that's going to be the average of the two numbers that are the endpoint.

02:12 So that will be the 66 .15.

02:16 On number nine, you want to find a 90 % confidence interval, and you know your sample mean is 55 .5.

02:24 0 .5, you have the t value with, for 90 % confidence, you'll have 19 degrees of freedom and 0 .05 in the upper tail, and that value corresponds to 1 .729, and then you'll take that sample standard deviation divided by the square root of 20.

02:45 For number 10 your confidence interval will be that 2 .45, i believe, is the sample mean.

02:57 You have 49 degrees of freedom, or excuse me, 48 degrees of freedom, and this is for a 95 % confidence interval.

03:05 So you want a t value with 0 .025 and 48.

03:09 I'm going to estimate this with using the 40 degrees of freedom.

03:13 That will be a pretty good estimate.

03:17 And that value is about 2 .021.

03:21 You can also use your t -cdf to find it more accurately.

03:24 And the sample standard deviation is 0 .25 and then divided by the square root of 49...

Question

Please give Ace some feedback