A random sample of 42 textbooks has a mean price of $114.50 and a standard devia
ID: 3127318 • Letter: A
Question
A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation of $12.30. Find a point estimate for the mean price of all textbooks. $42 $12.30 $114.50 $2.73 None Find the critical value z_c, necessary to form a 98% confidence interval. 2.33 2.05 0.5040 0.8365 None A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation of $12.30. Find a 98% confidence interval for the mean price of all textbooks. (111.38,117.62) (110.08,118.92) (110.78,118.22) (109.61,119.39) None Determine the minimum sample size needed to construct a 95% confidence interval for the mean age of employees at a company. The estimate must be accurate to within 0.5 year. Assume the standard deviation is 4.8 years. 18 19 354 355 None Find the critical value t_c necessary to form a 95% confidence interval with a sample size of 15. 1.960 2.145 2.131 2.120 NoneExplanation / Answer
5.
The best estimate is the sample mean,
OPTION C: $114.50 [ANSWER]
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6.
As c = 0.98, then
alpha/2 = (1-c)/2 = 0.01
Hence, using table/technology, the critical z is
OPTION A: 2.33 [ANSWER]
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7.
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.01
X = sample mean = 114.5
z(alpha/2) = critical z for the confidence interval = 2.326347874
s = sample standard deviation = 12.3
n = sample size = 42
Thus,
Margin of Error E = 4.415248223
Lower bound = 110.0847518
Upper bound = 118.9152482
Thus, the confidence interval is
( 110.0847518 , 118.9152482 ) [ANSWER, B]
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