Answered

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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

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