Answered

Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

QUESTION 2
A random sample of 254 college students were interviewed. about their study habits. One question asked was "How many hours do you study in a typical
week?". The sample had a mean of 23.4 hrs and a standard deviation of 7.6 hrs.
What would be the margin of error for a 95% confidence interval for the typical weekly study time of all college students from this sample?
Round your answer to two decimal places.

Sagot :

raphealnwobi

The margin of error for a 95% confidence interval is 0.93.

mean (μ) = 23.4, standard deviation (σ) = 7.6, sample (n) = 254, confidence = 95% = 0.95

α = 1 - C = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

The z score of α/2 is equal to the z score of 0.475 (0.5 - 0.025) which is equal to 1.96.

The margin of error (E) is given by:

[tex]E=z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} } \\\\E=1.96*\frac{7.6}{\sqrt{254} } \\\\E=0.93[/tex]

The margin of error for a 95% confidence interval is 0.93.

Find out more at: https://brainly.com/question/10501147

Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.