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In a survey of 80 college. students, students reported spending an average of 211 minutes per day on homework assignments (with 0 35) . Calculate a 95% confidence I...

Question

In a survey of 80 college. students, students reported spending an average of 211 minutes per day on homework assignments (with 0 35) . Calculate a 95% confidence Interval for the true mean time spent on homework assignments per day:

In a survey of 80 college. students, students reported spending an average of 211 minutes per day on homework assignments (with 0 35) . Calculate a 95% confidence Interval for the true mean time spent on homework assignments per day:



Answers

Constructing Confidence Intervals Use the data set to (a) find the sample mean, (b) find the sample standard deviation, and (c) construct a $99 \%$ confidence interval for the population mean. Assume the population is normally distributed. If convenient, use technology. The weekly time spent (in hours) on homework for 18 randomly selected high school students. $$\begin{array}{llrrrrrrr} 12.0 & 11.3 & 13.5 & 11.7 & 12.0 & 13.0 & 15.5 & 10.8 & 12.5 \\ 12.3 & 14.0 & 9.5 & 8.8 & 10.0 & 12.8 & 15.0 & 11.8 & 13.0 \end{array}$$

This exercise, we're going to be Constructing a 99.5 confidence interval for the meantime spent by all office workers in browsing the web in an eight hour day were given that the sample size, he is 50 The sample mean is 27.8 minutes and the sample standard deviation Is 8.2 minutes. And for us to calculate the 99.5% confidence interval, we shall substitute the values that we have here into the formula X bar plus or minus the critical valley for zet, multiplied by the sample standard deviation, since the population standard division is not known divided by the square of n. And when we substitute the values, we we obtain the following 27 0.8 plus or minus the critical value for the zet. That testing this and then there's, that score here Is going to be obtained as follows. So we have the 99.5% confidence interval which corresponds to the level of significance of 1 -0.995, Which equals 0.0 05. So the critical value that corresponds to this uh level of significance for a two tailed test is 2.807. And that's what we're going to substitute into the formula here. So it's 27.8 plus or -2.807 Times 8.2 Divided by the Square Root of 50. And when we saw that, when you work it out, it's going to be 27.8 plus or minus Uh 3.26 um that all that is given in minutes. So this is the 99.5% confidence interval for the population means. So we are 99.5% confidence confident that the population mean will be Within the interval given by 27, plus or minus that 3.26 minutes.

So we have this data given to us and actually we had a whole bunch of data, but we found out that that's 40 center deviation in my calculator was 1.24 and that's The mean. And I did all that just on my graphing calculator, we're looking for the 90's uh percent confidence interval. This is the formula we're going to use, we fill it in with all of this, and this is actually the Z score For a 90% confidence interval. And we're going to multiply that out, do plus and minus, and that is the 95% confidence interval that you should get. So.

Okay, so we want to come find a 95% confidence interval of day travelling distance for evening students and day student. So step one, we gonna gather. That's so didn't mean there stands for that day. Students is 4.7 something I don't remember if it smile. Army terrorists send their deviation for that populations that for that sample is 1.5 for the evening student debt sample. The simple is the mean for the simple evening student at 6.2 kilometers or mine I don't remember Miles I don't remember Center deviation for the population is 1.7 and both simple size are 40 individuals. So evening or day students 40 individual greater than 30. Although that's not really relevant to you. Now that we have over data, we're gonna find Z Alfa over to So it's Alfa is one minus 0.95 which is 5%. Therefore see off over two is Z 0.0 25 which is 1.96 Great. We have cf over to step three We compute z so compute Z First we remember that we take the mean of the days to evening students minus day student, the mean of samples. The difference of simple means is one point at 6.2 minus 4.7, which is one point by and now we compute Z Alfa over to a times distended deviation of each population, the variants of each population divided by their simple sigh. So 1.96 time Ah, 1.7 squared over 40 plus 1.5 square over 40 which is once you plug that in your capture, the chair c 0.7 and now we can compute our confidence interval. So find let's call this step four just for fun Spine. It's confidence Interval. So the difference in mean for day students or evening students is 1.5 minus 4.7 between my knees between my 1.5 minutes no 0.7 and 1.5 plus 4.7, which is zero point a and two points to So there we have it. We can have a tiny step Step five where we conclude So death 95% confidence interval for they mean the difference of that means of travelling distance for daytime and evening. Student is zero point A and 2.2. So quick recap we gathered Oliver data. We figured what was our ZF over to? We computed are we didn't have to compute. See, I don't know why I wrote that We come to you, Um, this thing here and you re computed Z alpha over to times this big thing in the square roof. And we used those computation to find their confidence interval, which is the report eight and 2.2.

The following is a solution in # five. And uh this asks a sample of 300 people if they work at home at least once a week and 35 of the 300 say that they do. So the number of favorable is 35. That'll be our X. And then the number of total would be 300. So if you wanted to find the p hat you can it's just 35 over 300. And we're asked to find the 99% confidence interval here. Now this one's a one proportions Z. Test so it's one prop Z. It's not test. I'm sorry interval. So one prop c. ent is this case because it's a proportion it's a favorable over total. We're looking at the percentage of people that work at home at least once once a week. Okay. So we're gonna do the 99% confidence and you can do this with the formula that may take you a little bit longer. I'm gonna use the T. I. 84 because it works pretty nicely. If you go to stat and then test and then go all the way down to the a option where it says one prop. Z ent All right. We got one proportion and we're just making the Z. And And this is where we get the x in the end. So the x was 35 And then the end was 300 And we want to be 99% confidence. We'll say 9 9. And whenever we calculate that's going to give us this top band here, that's the that's the confidence interval. And if you wanted you can have the P hat. That's about about 12% is the P hat. So let's go and write, write down that that interval there. So I'll go in around generally speaking I like to around 11 day, so 6.9% or .069. So I'll go to three decimal places here. Um so 0.692 point 164 with the numbers. So if you wanted to interpret that, we didn't have to because they didn't say, but if you wanted to interpret this, He would just say we we are 99% confident that the true proportion of people that work from home is between 6.9 and 16.4%.


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