12.1 Use the one-proportion z-interval procedure to find the required confidence
ID: 3324433 • Letter: 1
Question
12.1 Use the one-proportion z-interval procedure to find the required confidence intervall When 294 college students are randomly selected and surveyed, it is found that 115 own a car. 12) W hat is the best point estimate of the proportion of college students who own a car? Round to three decimal placess A) 0.391 B) 0.990 C) 2.557 D) 0.335 to 0.447 13) Find a 99% confidence interval for the proportion of all college students who own a car. A) 0.337 to 0.453 B) .318 to 464 C) 0.347 to 0.443 D) 0.326 to 0.463 14) What is the margin of error of the estimate? A) .146 B) .028 C).073 D).990 12.1 Assume that you wish to estimate a population proportion, p. For the given margin of error and confidence level, determine the sample size required. 15) You wish to estimate the proportion of shoppers that use credit cards. Obtain a sample size that will ensure a margin of error of at most 0.01 for a 95% confidence interval. In a previous survey, the percentage using credit cards was 70%. A) 13,925 B) 8068 C) 8067 D) 8067.36Explanation / Answer
TRADITIONAL METHOD
given that,
possibile chances (x)=115
sample size(n)=294
success rate ( p )= x/n = 0.3912
Q12.
sample proportion = 0.3912
standard error = Sqrt ( (0.3912*0.6088) /294) )
= 0.028
Q14.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
margin of error = 2.576 * 0.0285
= 0.073
Q13.
CI = [ p ± margin of error ]
confidence interval = [0.3912 ± 0.0733]
= [ 0.318 , 0.464]
Q15
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Sample Proportion = 0.7
ME = 0.01
n = ( 1.96 / 0.01 )^2 * 0.7*0.3
= 8067.36 ~ 8068
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