At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Connect with a community of experts ready to help you find solutions to your questions quickly and accurately. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Answer:
The 98% confidence interval for the mean amount spent on their child's last birthday gift is between $40.98 and $43.02.
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 = 24 - 1 = 23
98% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 23 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.98}{2} = 0.99[/tex]. So we have T = 2.5
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 2.5\frac{2}{\sqrt{24}} = 1.02[/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 42 - 1.02 = $40.98.
The upper end of the interval is the sample mean added to M. So it is 42 + 1.02 = $43.02.
The 98% confidence interval for the mean amount spent on their child's last birthday gift is between $40.98 and $43.02.
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
