6. A simple random sample of 25 internet users in Ada were...

6. A simple random sample of 25 internet users in Ada were asked how much data they downloaded during last month (in gigabytes, GB) and their answers are summarized as follows. The amount of data download is assumed to be normally distributed. $\bar{x}$=22.4; $\sum{x}$=560; $\sum{x^2}$=13508; $S$=6.3377; $\sigma$=6.2097; $n$=25 a. If we want to create a confidence interval for the average amount of data downloaded by all the internet users at Ada, shall we use a Z-interval or a t-interval? Justify your answer. b. Find the 99% confidence interval you have chosen in part a and interpret it in the context of this problem. c. What is the margin of error of this confidence interval? d. 5 If we calculate a 95% confidence interval instead, will the margin of error of this confidence interval be larger or smaller than the one in part c? State your reason. e. It is believed that the average amount of data download among Ohio internet users is 26 GB each month. Do the above data provide enough evidence that the average amount of monthly data download for Ada internet users is lower than state average? Conduct a test to answer this question. Make sure that you state the null and alternative hypotheses, find the value of the appropriate test statistic and its P-value. Draw your conclusion in the context of this problem at $\alpha$=0.01.

Transcript

00:01 We're going to suppose the weekly amount of time students spend in the university library is normally distributed.

00:06 Here's our normal curve.

00:08 And the standard deviation, sigma, is 1 .5 hours.

00:12 So we took a sample of 100 randomly selected students and found the mean time they spent in the library was 6 .5 hours.

00:20 And we're going to make a...

00:21 The first thing we're going to do is make a 95 % confidence interval for the population mean.

00:25 So 95 % confidence interval for mu.

00:31 And the formula is as follows.

00:33 We take our sample mean, x -bar, plus minus z alpha over 2.

00:39 And it's z because we know the population standard deviation is normally distributed.

00:43 So it's a z distribution.

00:46 And then the alpha, i'll just put that in here, 0 .05, because 1 minus the alpha gives us our confidence level.

00:52 And then we're going to multiply that by sigma over root n.

00:56 So we have everything we need except the z -score.

01:00 The z -score that corresponds with that alpha is 1 .96.

01:04 So the margin of error is 0 .294.

01:07 And that comes from the z multiplied by sigma over root n.

01:14 This is our margin of error.

01:16 So we're going to take 6 .5 plus or minus the margin of error.

01:20 And that's going to give us our interval.

01:23 So 6 .206.

01:29 So and you'll write...

01:30 You often see it written like this.

01:32 6 .206 up to 6 .794.

01:37 So that's a 95 % confidence interval for the population mean mu.

01:43 And so what this means is that we are 95 % confident that the true population mean time for a college student to spend in the library is contained in this interval.

01:57 It's not a probability statement.

01:59 It doesn't mean there's a 95 % chance it's in there.

02:01 It's that we're 95 % confident we've captured it.

02:04 And let's say we were to bump this up to a 99 % confidence interval.

02:09 Well, this, without doing any calculations, the way we become more confident is we make our interval wider.

02:14 So we would go outside this round.

02:17 So we'd make it a little bigger.

02:18 I'll do it in a separate color.

02:19 Let's say it's this one.

02:20 We'd get a little...

02:21 We'd extend the confidence interval a little bit more.

02:24 If you're thinking about these values on a number line, so this is a number line.

02:38 Okay, two values here.

02:41 To make it...

02:41 To be more confident, what you do is you extend your interval.

02:44 You make it right bigger.

02:45 And then what you do, this alpha would change, 0 .01.

02:50 And then this z -score is going to get a little bit bigger.

02:53 And that's why it's going to get wider.

02:57 So now we're going to say we have a staff member who believes that students don't spend any more than two and a quarter hours in the library.

03:06 So our hypotheses for this statement, for this claim, would be that, all right, the mean is less than or equal to 6 .25 hours.

03:16 The alternative hypothesis is that the mean is in fact greater than 6 .25 hours.

03:22 Because the librarian's claiming, no, they don't spend...

03:24 They spend no more than six and a quarter hours.

03:28 But then the alternative would be, well, they could spend more than six and a half.

03:32 And so here we see the mean is six and a half.

03:36 We're going to say, hey, is this six and a half significantly greater than six and a quarter? and this is going to be a one -tailed test because we're looking strictly greater than.

03:46 So here's our mean, which corresponds to a z -score of zero.

03:51 And then we have some value up here, 6 .5, which is going to correspond to some z -value here.

03:59 And then what we're looking for is this 6 .5 value, is it significantly greater than? and actually, let me scale this differently.

04:07 Because we're going to test this at the alpha of 0 .10 level of significance, which means that we're going to have some critical value, i should say.

04:23 This is where i should have some...

04:24 We have some z -critical value, which corresponds to some critical x -value, x -bar value, such that the error to the right here is 0 .10.

04:41 And we're going to see, is this 6 .5, does that fall in here? so we're going to convert that 6 .5 into a z -score with the following formula.

04:49 Z is x -bar minus the mean of the sampling distribution divided by sigma over root n.

04:56 And so we have everything we need, 6 .5, 6 .25, and sigma we know, 1 .5 and n is 100.

05:04 Let's go ahead and do that.

05:08 And this is what we get.

05:10 So here we go...

Question

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