Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
Using the z-distribution, it is found that the 90% confidence interval for mean calories in a 30-gram serving of all chocolate chip cookies is (143, 149).
We are given the standard deviation for the population, which is why the z-distribution is used to solve this question.
The information given is:
- Sample mean of [tex]\overline{x} = 146[/tex].
- Population standard deviation of [tex]\sigma = 12[/tex].
- Sample size of [tex]n = 65[/tex].
The confidence interval is:
[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]
The critical value, using a z-distribution calculator, for a 95% confidence interval is z = 1.645, hence:
[tex]\overline{x} - z\frac{\sigma}{\sqrt{n}} = 146 - 1.645\frac{12}{\sqrt{35}} = 143[/tex]
[tex]\overline{x} + z\frac{\sigma}{\sqrt{n}} = 146 + 1.645\frac{12}{\sqrt{35}} = 149[/tex]
The 90% confidence interval for mean calories in a 30-gram serving of all chocolate chip cookies is (143, 149).
A similar problem is given at https://brainly.com/question/16807970
We hope our answers were helpful. Return anytime for more information and answers to any other 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 using Westonci.ca. Come back for more in-depth answers to all your queries.
