Question

Score: 0 of 1 pt 7 of 20 (1 complete) 7.1.15 Use the sample data and confidence level given below to complete parts (a) through (d). In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2432 subjects randomly selected from an online g proportion of returned surveys. 2 Click the icon to view a table of z scores. a) Find the best point estimate of the population proportion p. (Round to three decimal places as needed.) Enter your answer in the answer box and then click Check Answer. 3 remaining parts Clear All

Answer 1

Solution

given: 0 919, n = 2432

Sample prportion () = 919 2432 = 0.378

(a) Point estimate for p = p = 0.378

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(b) for 99% confidence level, ze 2.576

Confidence level (C) = 0.99 a = 1-C=1-0.99 0.99 = 0.01 a = 0.01 = 0.005 2 2 Refer Standard normal table/Z-table or use excel function "=NORM.S.INV((1-0.005))" to find the critical value of z. -.24 = 2.576

p* (1-P) Margin of error (E * n

$\small \mathbf{Margin\;of\;error\;(E)}= 2.576*\left \{ \sqrt{\frac{0.378*(1-0.378)}{2432 }} \right \}$

Margin of error E) = 0.025

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(c

formula : Confidence interval for the population proportion, p

= P + Margin of error E

$=0.378 \pm 0.025$

$\mathbf{lower\;limit}=0.378- 0.025=\mathbf{{\color{Red} }0.353 }$

$\mathbf{upper\;limit}=0.378- 0.025=\mathbf{{\color{Red} }0.403 }$

$\mathbf{\therefore \mathbf{{\color{Red} }{\color{Red} 0.353}} <p<\mathbf{{\color{Red} }{\color{Red} 0.403}} }$

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$\mathbf{{\color{Blue} (d)}}$  One has 99% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.