The following data represent the repair cost for a low-impact collision in a simple random sample f mini- and micro-vehicles_ 53148 S1758 S1071 S3345 5743 S20...

Repair Costs: Refrigerators In a random sample of 60 refrigerators, the mean repair cost was $\$ 150.00 .$ Assume the population standard deviation is $\$ 15.50 .$ Construct a $99 \%$ confidence interval for the population mean repair cost. Interpret the results.

So in this problem, we're being asked to construct a 95% confidence interval based off of the given information. So the first thing we want to do in this situation since we're working with the T distribution is fine. The T value. Once we find the t value, we can use that to find the margin of error and ultimately to find the confidence interval when faced was the word problem like this. The first thing we want to do is look through the problem and try and find all of the relevant information. So, in this problem, by doing a simple read through, we have computer workers. So in a random sample of seven computers, so we know n equals seven. The main repair costs was $110 so x bar is 110 and the standard edition was 44.5 dollar 44 50 which is 44.5. And we know that the confidence level from the problem is 0.95 So first, to find the T value, we need to determine the degrees of freedom which is equal to sample size minus 17 minus one or six, then we already know that the confidence level is 0.95 So if we look in the table in table five of Appendix B, we see that this is equal to 2.447 Now, given this t level T value, we know that the margin of error is equal to teeth times Yes, over the square root of em plugging in our values. This is 2.447 times 44.5 over the square root of seven. And this is approximately equal to 41 0.157 Now we simply we know the confidence interval is simply from X bar minus the margin of error. Thank you to the mean X bar, plus the margin of error. So plugging in our values, this would be 1 10 minus 41.157 just equal to 68.843 toe 1 10 plus 41.157 which is equal to 1. 51.157 Okay. And so this gives us the confidence interval of 68.843 toe 1. 51.157 and that's a final answer, and we also know that the margin of error is 41.157

So in this problem, we're being asked to construct a confidence interval using the given information in a word problem. Now first, just to recap when constructing confidence Interval using a T distribution, which is when we do not know the sample standard. When we do not know the population standard deviation sigma we first need to find the TC value and once we do that, we can use it to find the margin of error and finally use that to find our confidence interval. Now, when given a word problem, the first thing we need to do is read, read through and extract all the relevant information. So just to read through. So we have microwave repairs in a random sample of 13 microwave ovens. So from that line, we know the end equals 13. Sample size is 13. The main repair costs was $80 so the mean is $80 and the standard deviation was 13 50. And so this is the sample standard deviation and its 13 50 now for the confidence level. We know from the question that we're trying to find a 95% confidence interval, so we will set it at 950.95 Now, when given a set up like this, the first thing we need to do is determine the T T value, which weaken do by knowing the confidence level as well as the degrees of freedom. And the degrees of freedom is simply the sample size minus one, which is 13 minus one, which is 12. Now flip the table five in your appendix B, and you can see that the T value for 12 degrees of freedom at 120.95 confidence level is 2.179 Now, to find the margin of error, we know that the margin of error is teeth. Times s over the square root of n and plugging in our values. That is 2.179 times 13 50 over square root of 13. And that's equal to approximately 8.159 Now. To find our confidence interval, we know our confidence. Interval is the average minus that margin of error to the average, plus the margin of error. So to find those values, we have 80 minus 8.159 which is equal to 71.841 to 80 plus 8.159 which is equal to 88.159 And there you go, we have a confidence interval of 71 0.841 Thio 88.159 And that's your final answer. And we also know that our margin of error is 8.159

So we are given this data atop were asked to use it to find the 95% confidence interval, and this is our confidence interval formula. While the average goes there, The Z score for a 95% confidence interval, I looked it up in the table is that you multiply it by the standard deviation, divided by square defend. And when you put all that in your calculator and you do a positive and a negative or a plus and a minus, this is the interval that you should get yeah.

Now, in this case, we have randomly selected 50 replacement costs, which means R N s 50 r is equal to 50. That is, our sample size is 50 and we find that our meanness to $650 are mean. Or I can say expert for the sample is to $650. We are given The population center division has 4. 25 Sigma is equal to 4. 20 life and we want a 95% confidence interval. So if I use and online dude to calculate my confidence in trouble or I can also use this formula X bar plus minus zed Alfa by two Sigma by 10. Okay, this is going to give me lower and upper limits for my confidence interval. Okay. What is going to be my alphabet to my Alfa 0.0 fights in my alphabet to will become my alphabet will become 0.0 to fight. But Okay, well, do this using a calculator. So if I put these values in the calculator and I hit calculate This is giving me 2650 plus minus 117.8. This is 26 50 plus minus 117.8. So if I write, then double. This is 2650 less 11 seven. So this is 52533 or 2532 rather to five 32.2. Common. This is 2767.8. So this is my confidence interval. So I am 95% confident that my meaning replacement cost off to mean replacement cost off. The population is going to lie in this region, and this is my answer.