5. A random sample of 84 shoppers were interviewed and 51 said they prefer to sh
ID: 3133988 • Letter: 5
Question
5. A random sample of 84 shoppers were interviewed and 51 said they prefer to shop alone
rather than with someone such as friends or family.
A. Let p represent the proportion of all shoppers at this mall who would prefer to
shop alone. Find a point estimate p for p. A. _______
(a) 0.393 (b) 51 (c) 84 (d) 0.607 (e) 0.5
B. Find a 90% confidence interval for p. B. _______
(a) 0.519 to 0.695 (b) 0.517 to 0.697
(c) 0.20 to 1.41 (d) 0.503 to 0.711 (e) 0.305 to 0.481
C. How many more students should be included in the sample to be 90% sure that
a point estimate p will be within a distance of 0.05 from p. C. _______
(a) 271 (b) 187 (c) 259 (d) 283 (e) 175
Explanation / Answer
5.
A)
Note that
p^ = point estimate of the population proportion = x / n = 0.607142857 = 0.607 [ANSWER, D]
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b)
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.053287228
Now, for the critical z,
alpha/2 = 0.05
Thus, z(alpha/2) = 1.644853627
Thus,
Margin of error = z(alpha/2)*sp = 0.08764969
lower bound = p^ - z(alpha/2) * sp = 0.519493167
upper bound = p^ + z(alpha/2) * sp = 0.694792547
Thus, the confidence interval is
( 0.519493167 , 0.694792547 ) [ANSWER, A]
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c)
Note that
n = z(alpha/2)^2 p (1 - p) / E^2
where
alpha/2 = 0.05
Using a table/technology,
z(alpha/2) = 1.644853627
Also,
E = 0.05
p = 0.607142857
Thus,
n = 258.1309316
Rounding up,
n = 259 [ANSWER, c]
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