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Answer :
Answer:
We are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%
Step-by-step explanation:
We are given that in a group of randomly selected adults, 160 identified themselves as executives.
n = 160
Also we are given that 42 of executives preferred trucks.
So the proportion of executives who prefer trucks is given by
p = 42/160
p = 0.2625
We are asked to find the 95% confidence interval for the percent of executives who prefer trucks.
We can use normal distribution for this problem if the following conditions are satisfied.
n×p ≥ 10
160×0.2625 ≥ 10
42 ≥ 10 (satisfied)
n×(1 - p) ≥ 10
160×(1 - 0.2625) ≥ 10
118 ≥ 10 (satisfied)
The required confidence interval is given by
[tex]$ p \pm z\times \sqrt{\frac{p(1-p)}{n} } $[/tex]
Where p is the proportion of executives who prefer trucks, n is the number of executives and z is the z-score corresponding to the confidence level of 95%.
Form the z-table, the z-score corresponding to the confidence level of 95% is 1.96
[tex]$ p \pm z\times \sqrt{\frac{p(1-p)}{n} } $[/tex]
[tex]$ 0.2625 \pm 1.96\times \sqrt{\frac{0.2625(1-0.2625)}{160} } $[/tex]
[tex]$ 0.2625 \pm 1.96\times 0.03478 $[/tex]
[tex]$ 0.2625 \pm 0.06816 $[/tex]
[tex]0.2625 - 0.06816, \: 0.2625 + 0.06816[/tex]
[tex](0.1943, \: 0.3306)[/tex]
[tex](19.43\%, \: 33.06\%)[/tex]
Therefore, we are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%
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