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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 None

Explanation / 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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