Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
Answer: 8.323 (approximate)
====================================================
Explanation:
s = 32 = sample standard deviation
n = 40 = sample size
Despite not knowing the population standard deviation (sigma), we can still use the Z distribution because n > 30. When n is this large, the student T distribution is approximately the same (more or less) compared to the standard Z distribution. The Z distribution is nicer to work with.
At 90% confidence, the z critical value is roughly z = 1.645. This can be found using either a Z table or a calculator.
------------------
We have these values:
- z = 1.645 (approximate)
- s = 32
- n = 40
Plug these values into the margin of error formula.
[tex]E = z*\frac{s}{\sqrt{n}}\\\\E \approx 1.645*\frac{32}{\sqrt{40}}\\\\E \approx 8.32311480156318\\\\E \approx 8.323\\\\[/tex]
The margin of error is roughly 8.323
I rounded to 3 decimal places because z = 1.645 is rounded to 3 decimal places. If your teacher wants some other level of precision, then be sure to follow those instructions.
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
