Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
Using the t-distribution, as we have the standard deviation for the sample, it is found that the 95% confidence interval for the number of units students in their college are enrolled in is (11.7, 12.7).
What is a t-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
In which:
- [tex]\overline{x}[/tex] is the sample mean.
- t is the critical value.
- n is the sample size.
- s is the standard deviation for the sample.
The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 49 - 1 = 48 df, is t = 2.0106.
Hence:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 12.2 - 2.0106\frac{1.6}{\sqrt{49}} = 11.7[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 12.2 + 2.0106\frac{1.6}{\sqrt{49}} = 12.7[/tex]
The 95% confidence interval for the number of units students in their college are enrolled in is (11.7, 12.7).
More can be learned about the t-distribution at https://brainly.com/question/16162795
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
