Answered

IDNLearn.com: Your trusted platform for finding precise and reliable answers. Our community is here to provide the comprehensive and accurate answers you need to make informed decisions.

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 :

Raphealnwobiidn

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

Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.