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Week working at home after school. Construct and interpret a 95% confidence interval for the mean number of hours per week a teacher spends working at home. Assume ...

Question

Week working at home after school. Construct and interpret a 95% confidence interval for the mean number of hours per week a teacher spends working at home. Assume that the population standard deviation is 1.5 hours per week.

week working at home after school. Construct and interpret a 95% confidence interval for the mean number of hours per week a teacher spends working at home. Assume that the population standard deviation is 1.5 hours per week.



Answers

According to the Bureau of Labor Statistics publication News, self-employed persons with home-based businesses work a mean of 25.4 hours per week at home. Assume a standard deviation of 10 hours. a. Identify the population and variable. b. For samples of size $100,$ find the mean and standard deviation of all possible sample mean hours worked per week at home. c. Repeat part (b) for samples of size $1000 .$

Yeah, We obtain a random sample with size and equals 11195 and with sample mean expert equals 9.2, Assuming that the population of the standard deviation 6.7 there's no shock to 90% confidence interval from you. This question is testing an understanding of how to test that or estimated parameter in particular. Testing us on how to construct a confidence interval for hypothesis from you. So we proceed the steps A through C. To solve first we calculate the margin of error of our interval. This is E equals Z alpha over two times sigma overwritten Z alpha over to depend on the confidence level. So for 90% confidence from a normal curves calculator we get C alpha over two equals 1.645 Plugging in the three parameters gives equals 30.32 Next we construct the confidence interval as X, Y plus or minus E. The confidence interval is experiment etc. is 8.8 two x bar plus is 9.52. Finally, we conclude that we are 90% confident. The mean is between our lower and upper bound cited in the confidence interval.

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%.

Following a solution for number 16 and this study's theme park spending. So there was a random sample of 40 visitors and the mean amount spent per person per day at that particular theme park is $93.43. That's a lot. And it says the population standard deviation is 15, so we actually have sigma here, the population standard deviation sigma is 15 and we're asked to find the 95% confidence interval for the mean amount spent per person per day at this particular um theme park. So what we're gonna do here is determine which method to use and it's going to be the Z interval. The reason why it's a Z interval. Well one, we're testing, not testing, we're estimating the mean, so it's got to be there the tear the Z interval and were able to use the Z interval because we actually know what sigma is. So um we know the population standard deviation. We can be a little bit more accurate with that Z interval. So you can use the z interval formula or you can use technology. I'm gonna use technology because it goes by a little bit more quickly. But if you go to stat and then test, it's the seventh option here is the interval And just make sure summary stats is highlighted there. And remember sigma, the population standard deviation was 15, The x bar was 93.43 And the sample size was 40 visitors and we want to be 95% confidence of 95 is the sea level. And then whenever we calculate this top band here, that gives us the confidence interval. So 88.78 and 98.08, I'm rounding to the nearest sent here, so I'm rounded to two decimal places. So 88 point 78 and 98 08 So a little dollar signs there. Okay, so that's my confidence interval now, I actually have to interpret that confidence interval with um with the theme at hand. So here's how these confidence intervals, they kind of follow the same pattern. But we're gonna say We are 95% confident I'm typing it because it'll be neater this way that the mean amount of money all people spend daily per person at this theme park Is between 88. Just put dollar in front of it, 88 points mm 0.78 and $98.08. Okay, so we're 95% confident that the mean amount of money that all people spend daily per person at this theme park is between $88.78.98.08.

In this problem. We given that a random example off 15 NASA's in a large hospital should that they worked on average off 44.2 hours a week. And it also given that the standard division of example was 2.6. And with that information, we're going to be estimating the mean off the population with 90% confidence. So we start off with the expression that will give us the 90% confidence interval and that is given by X bar minus t multiplied by started division. If I had by the scarred off n less than volition mean yeah, she's less than X bar plus t multiplied by s squared off in now the value off T when for our case here eyes going to be determined by the sample size and the sample size and is 15. So the value off t at the 90% confidence obtain from the graph from the table F is equal to one point 761 because we have the degrees of freedom as n minus one, which in this case is 15 minus one, which is 14. So when you check, you'll find that the value of tea is 1.7 61 Now the Standard Division s according to what we have been given is 2.6. So now we can substitute the values into the expression and that will give us the 90% confidence interval. So in this case will have our X bar, which is 44.2 minus 1.761 multiplied by 2.6 divided by the square it off 15. That's less than pleasure. Means you just less than X bar. I'm sorry. In this case, we have 44 0.2 plus 1.761 multiplied by 2.60. I was carried off 15. Now, now this would be 44.2 minus. When you solve this whole partier, you're going to team Avandia 1.18 Then proceed. Get 4.44 and to plus 1.18 Now, when we simplify, you're going to get 14 three 0.2 It's less than view, which is less than 45 38 and this is the estimate off the main off the population with 90% confidence


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