IDNLearn.com offers a reliable platform for finding accurate and timely answers. Get accurate and detailed answers to your questions from our dedicated community members who are always ready to help.
Sagot :
Answer:
The 98% confidence interval for the mean usage in the March quarter of 2006, in kWh, was (333.87, 416.13).
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 30 - 1 = 29
98% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 29 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.98}{2} = 0.99[/tex]. So we have T = 2.462
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 2.462\frac{91.5}{\sqrt{30}} = 41.13[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 375 - 41.13 = 333.87 kWh
The upper end of the interval is the sample mean added to M. So it is 375 + 41.13 = 416.13 kWh
The 98% confidence interval for the mean usage in the March quarter of 2006, in kWh, was (333.87, 416.13).
Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. Thank you for choosing IDNLearn.com for your queries. We’re committed to providing accurate answers, so visit us again soon.
