A random sample of 15 DVD players has a mean price of $64.30 and a standard devi
ID: 3127321 • Letter: A
Question
A random sample of 15 DVD players has a mean price of $64.30 and a standard deviation of $5.60. Find a 95% confidence interval for the mean price of all DVD players. (61.20, 67.40) (61.47, 67.13) (61.22, 67.38) (61.10, 67.51) None In a survey of 250 Internet users, 195 have high-speed Internet access at home. Find a point estimate for the proportion of all internet users who have high-speed internet access at home. 1.28 0.78 0.22 195 None In a survey of 250 Internet users, 195 have high-speed Internet access at home. Find a 90% confidence interval for the proportion of all Internet users who have high-speed Internet access at home. (1.19, 1.38) (0.728, 0.832) (0.731, 0.829) (0.737, 0.823) None You want to estimate, with 95% confidence, the proportion of households with pets. Your estimate must be accurate within 3% of the population proportion. No preliminary estimate is available. Find the minimum sample size needed. 1141 3267 1068 1067 NoneExplanation / Answer
10.
Note that
Margin of Error E = t(alpha/2) * s / sqrt(n)
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 64.3
t(alpha/2) = critical t for the confidence interval = 2.144786688
s = sample standard deviation = 5.6
n = sample size = 15
df = n - 1 = 14
Thus,
Margin of Error E = 3.101176633
Lower bound = 61.19882337
Upper bound = 67.40117663
Thus, the confidence interval is
( 61.19882337 , 67.40117663 ) [ANSWER, A]
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