Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A 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 appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
