At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
Using the t-distribution, it is found that the 95% confidence interval for the mean number of people the houses were shown is (20.1, 27.9).
We have the standard deviation for the sample, hence the t-distribution is used to build the confidence interval. Important information are given by:
- Sample mean of [tex]\overline{x} = 24[/tex].
- Sample standard deviation of [tex]s = 9[/tex].
- Sample size of [tex]n = 23[/tex]
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
In which t is the critical value for a 95% confidence interval with 23 - 1 = 22 df, thus, looking at a calculator or at the t-table, it is found that t = 2.0739.
Then:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 24 - 2.0739\frac{9}{\sqrt{23}} = 20.1[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 24 + 2.0739\frac{9}{\sqrt{23}} = 27.9[/tex]
The 95% confidence interval for the mean number of people the houses were shown is (20.1, 27.9).
A similar problem is given at https://brainly.com/question/15180581
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
