In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2457 subjects randomly selected from an onl

Question

In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2457 subjects randomly selected from an online group involved with ears. 935 surveys were returned. Construct a 95% confidence interval for the proportion of returned surveys.Find the best point estimate of the population proportion p.Identify the value of the margin of error E.Construct the confidence intreval.

in progress 0
Mary 1 week 2021-10-09T19:49:01+00:00 1 Answer 0

Answers ( )

    0
    2021-10-09T19:50:06+00:00

    Answer: a) Margin of error = 0.19, b) interval = (0.19,0.57)

    Step-by-step explanation:

    Since we have given that

    n = 2457

    x = 935

    So, p=\dfrac{935}{2457}=0.38

    At 95% confidence , z = 1.96

    First we will find margin of error.

    So, Margin of error is given by

    z\sqrt{\dfrac{p(1-p)}{n}}\\\\=1.96\times \sqrt{\dfrac{0.38\times 0.62}{2457}}\\\\=0.19

    95% confidence interval is given by

    p\pm \text{margin of error}\\\\=0.38\pm 0.19\\\\=(0.38-0.19,0.38+0.19)\\\\=(0.19,0.57)

    Hence, a) Margin of error = 0.19, b) interval = (0.19,0.57)

Leave an answer

Browse
Browse

27:3+15-4x7+3-1=? ( )