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It is claimed that the mcan age of a BCIT night school student is 30 years old: You collect random samplc of 300 BCIT night schoon students and construct 99% confid...

Question

It is claimed that the mcan age of a BCIT night school student is 30 years old: You collect random samplc of 300 BCIT night schoon students and construct 99% confidence interval for the mean agc.27 _ 6 ycars @99%Does this confidence interval rcject the claim that the mcan age of BCIT night school students is 30 years old? Explain: Do not perform forma hypothesis testNo thc intervab does not rcject thc claim that the mcan agc is 30 ycars old sincc 30 is within the confidence interval.Yes; the int

It is claimed that the mcan age of a BCIT night school student is 30 years old: You collect random samplc of 300 BCIT night schoon students and construct 99% confidence interval for the mean agc. 27 _ 6 ycars @99% Does this confidence interval rcject the claim that the mcan age of BCIT night school students is 30 years old? Explain: Do not perform forma hypothesis test No thc intervab does not rcject thc claim that the mcan agc is 30 ycars old sincc 30 is within the confidence interval. Yes; the interval docs reject the claim that the mcan agc is 30 years old since thcy are only 99% confident: Yes_ the interva docs rcject the claim that the mcan agc is 30 ycars old since thc mean agc of the sample was much lower; 27-ycars-old Yes; the interval docs reject the claim that the mcan agc is 30 years old since 30 is outside the confidence intcrval:



Answers

Explaining confidence The admissions director from Big City University found that $(107.8,116.2)$ is a
95$\%$ confidence interval for the mean 1$Q$ score of all freshmen. Comment on whether or not each of the following explanations is correct.
(a) There is a 95$\%$ probability that the interval from 107.8 to 116.2 contains $\mu .$
(b) There is a 95$\%$ chance that the interval $(107.8,$ 116.2 ) contains $\overline{x}$ .
(c) This interval was constructed using a method that produces intervals that capture the true mean in 95$\%$ of all possible samples.
(d) 9$\%$ of all possible samples will contain the interval $(107.8,116.2) .$
(e) The probability that the interval $(107.8,116.2)$ captures $\mu$ is either 0 or $1,$ but we don't know which.

Part one. This is the result of the simple regression equation. The estimate on education is 0.1 oh one with a standard barrel of 10.7 This variable is highly significant, and you can calculate the 95% confidence interval of the return to education as a range of from 8.7% to 11.5%. Recall that the 95% confidence interval is calculated as the estimate plus or minus the margin of error and the margin of error is the standard error times a critical value. We need to look at the alpha level of 5% because we need the 95% confidence. Yeah, you should have the alpha level because this is a two sided interval. The degree of freedom is the number of observations minus two. We need to degree of freedom or two pieces of information to estimate the slope. And there, uh, the to intercept the intercept and the slope coefficient. So this is what you get. Part two. The simple regression of education on C two. It gives the variable see to it is not significant. The T statistic, which is the ratio of the estimate over the standard error is minus 0.59 And so we cannot use see to it as an instrument for education. Part three. The multiple regression equation estimated by L s is as follows. The coefficient on education is 1.137 experience 0.11 to experience square point oh three. And we have other factors which are not our main focus in this problem. Mhm, even the estimate on education. You can see that the estimated return to education is now higher 13.7%. We convert the estimate on education 2% because our dependent variable, the left hand side variable is in log and education is in level Part four in the multiple regression above. In part two. Mm hmm. No. The one with other explanatory variables. In part three, the coefficient on C two. It is minus point 165 Yeah, and the T statistic is minus 2.8. So an increase of $1000? Yeah. In institution reduces years of education by about Hong 165 Remember that the Tunisian variables are measured in thousands. Part five. Now we estimate the multiple regression model by Ivy using See to it as an I V for education. The I V estimate of beta education is point 25 with a standard error of 250.1 to 2. The point estimate seems large, but the 95% confidence interval is very white from 1.1% to 48.9%. We could reject the value zero for beta education, but this confidence interval is too white to be useful. Lastly, part six very large standard error of the Ivy Estimate in Part five shows that the ivy analysis is not very useful. Yeah, and see to it is not convincing as an I V, as we can see also from part two mhm.

So we would be assuming that are null hypothesis is that the mean amount of time after taking these open call Arklow and I don't remember what it's called, Let's just call it drugs E for insomnia, that the meantime is 102.8 minutes, the same as what it was before taking the drug. And alternately their insomnia decreases so that they're awakened. Time is less than that. And we have a sample size of 16. And from those 16 people they had a mean time of 98.9 minutes, with a standard deviation that's pretty big, 42.3 minutes. That's a very large standard deviation compared to that mean. So we would be assuming that the mean amount of time that they are awake is that 102.8 and that we're getting a 98.9. But we're going to assume it's, yeah the mean is actually this and we want to find that this is R. P. Value. And we need to find the test statistic that corresponds with that and it's a. T. Value. So how likely is it if the mean is truly this amount, how likely is it to sample from that distribution and to get a mean? That is less than or equal to 98.9. Now let's find what that test statistic is with 15 degrees of freedom and we find that by taking the 98.9 minus the one oh 2.8 divided by the standard deviation over the square root of em. And that test statistic comes out to be and let's get that we have that 98.9 minus and I'm putting a left parentheses and first I could've subtract that in my head but you know, it's late in the day and didn't feel like doing it. So I put the numerator in with parentheses and left and right and then my denominator left parenthesis. E 42.3 divided by let's just say four. Since that we can do that math in our head. And when we do we get that test statistic is only negative 0.3688 That's a six. So there's the test statistic and now we're going to find that P. Value and I'm going to use my T. C. D. F. Instead of looking it up in a table. And I'm going to put the left number is like negative 1000 the right number negative 10000.3688 and my degrees of freedom of 15. And now I'm actually gonna type it in because I couldn't say it and type it in at the same time. So negative 1000 and the I could just use second answer to get that value in there. Got 15 degrees of freedom paste. And we got it. We got our T. Value. Excuse me. R. P. Value is 0.3587 So this area is about 36%. So we have fail to reject. Yeah the National in other words this you pick any significance level. There is no way that you're going to have a significance level. That's like 50% significance level. If you tip typically significance levels 1% 2% 3%. 4% 5%. Maybe 10%. Well this is way larger than those. So this is normal variability if the mean is actually 102.8 minutes. So we failed to reject the null. So it seems like after the drug after the drug after the Z. Uh It appears that it is not effective not effective because we don't appear to get results that are any different, significantly different. Yes, it's lower, but it's not considered to be significantly lower, so it looks like the drug is not working.

Mhm. This is a template off one of the past questions that we did. We're going to use this same template again. This all this question as well. Let's go through the question. Ah, clinical trial was conducted to test the effectiveness off a drug Zopa clone for treating insomnia in older subjects. Before treatment with Rubicon, 16 subjects had a mean time off. 102.8. Okay, So mean off 102.8. All right. After the treatment, the 16 subjects had a mean wake time off. 98.9. Afterwards this was before and x bar. This will be after afterwards. What did they have? 98.9 98.9. Okay, under standard deviation off 43.2, 43.2 minutes and s waas 43.2 minutes. All right. Assume that the 16 sample values appeared to form Normally distributed population can construct in 98% confidence interval for the estimate off. Mean weight over the population with Zo pick long treatments with topical treatments. That is after they are treated. So this is going to be my I mean, this is going to be my sample standard deviation and in this case is equal to 16. So degree of freedom is going to be degree off. Freedom is going to be 15 now. I want a 98% confidence interval, which means my Alfa is equal to 0.2 So what is going to be my Alfa by two Alfa by 20.1 And what is the formula that we're using? It is this very same formula expert plus minus the margin of error and margin of a resident by the Alfa by two as by routine. And when I do this, the confidence interval that I'm getting is 74. Sorry. 71.421 26.4. 71.4 21 26.4. 71. Excuse me. 71.421 26.4. This is the confidence interval that I'm getting. What does the results suggest? About the mean time of 10 to before the treatment? Well, what was the treatment? Uh, to test the effectiveness for treating insomnia. Okay, so 102 does appear to fall in this region. So we can say that there is not enough statistical evidence to conclude anything because it does fall in this region. So we do not have enough evidence to say that the sleep time has actually increased or decreased. We cannot say anything so the drug does not seem to be very effective or does not seem toe change mute by much before New Waas 102 pointed and after that newest 98.9. So that is it does not seem to be very effective.

Hello. Hi. Hearing this question, we have to determine the chi square value first. Okay, So the critical value at 98% of confidence, we had to find out on when you Skype sky skirt table first V c. A number off samples given here is 49. Okay, so you can find a decrease of freedom. Degrees of freedom will be getting and minus one that is equal to 48. You can see that 48 value is not there in the table on you can see 48 is greater than 40 from The terrible will be using the value 40 in order to get the values off particular values. Okay, now see values 98% so we can write physical 2.98 Then you can obtain Alfa values. I'll physical 21 minus e one minus three is one minus 10.981 minus 0.98 You can write 0.0 toe, then left. Detailed critical value. Left tail critical value that is chi square one minus alphabet will be K square one minus. Alfa by. To hear, it will be 0.99 from the table. We get it is the call to 22 point 164 Next to write tailed critical value. This guy square alphabet. Do you get square off? 0.1 from the table. We obtained the value 63.691 Okay, once we get chi square values, the next thing we can find out, we can find the confidence in trouble. Estimate four Standard deviation. We know the formula. The formula is given by what is square root off and minus one by Chi Square Alfa by two times standard deviation less than sigma. Less than squared off and minus one by chi. Squared one minus Alfa by two times standard deviation. Okay, so once we know these values, we just start substitute all these values given and minus one value. You know that is equal to 48 Chi square. Well, we found out that is equal toe 63 point on 691 times. The standard deviation value we know it is given in the question is is equal to 21. We get next to less than Sigma less than here on the right. Answer with how again? 48 by the chi square value. We know that is able to convict 2.164 times the standard deviation again. Two D one. So we just started simplified these values. Okay, when you simply find you get the value 18 point 2306 less than standard deviation value less than 13 point 9041 Okay, so from here, you can see this value does not give it any information about effectiveness. Right? You can see this value does not give any information about them effectiveness. While the value we can still stigma values between 18.23 on 13.9. Right? So that's why we cannot have any information about effectiveness off particular value. Here. I hope that this answer your question. Thank you.


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