A marketing research company hired by Pepsi is interested in estimating the proportion of cola customers in Ontario who prefer Pepsi to other brands. A simple random sample of 120 Ontario cola customers was selected and 45% of them reported favoring other brands over Pepsi.
What is the upper limit of the large sample 95% confidence interval for the true proportion of cola customers in Ontario who prefer Pepsi to other brands?
(Hint – use 6 decimal places for intermediate calculations to avoid too much rounding error.)
A. 0.639
B. 0.089
C. 0.088
D. 0.636
E. 0.539
Solution :
Given that,
n = 120
Point estimate = sample proportion = =45% = 0.45
1 - = 1- 0.45 = 0.55
Z_{/2} = 1.960
Margin of error = E = Z_{ / 2} * (( * (1 - )) / n)
= 1.960 * ((0.45*(0.55) /120 )
= 0.089
A 95% confidence interval for population proportion p is ,
- E < p < + E
0.450- 0.089 < p < 0.450+0.089
0.361< p < 0.539
The 95% confidence interval for the population proportion p is : 0.361,0.539
Upper limit = 0.539
E. 0.539
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