A market researcher for a consumer electronics company want

A market researcher for a consumer electronics company want to study the television viewing habits of residents of a particular area. A random sample of 40 respondents is selected, and each respondent is instructed to keep a detailed record of all television viewing in a particular week. The results of the sample indicate the average viewing time is 15.3 hours and the standard deviation is 3.8 hours. In this scenario, 99% confidence interval estimate of the population mean has a margin of error of 1.55. If the sample size were decreased to 30, the margin of error would get smaller and the width of the interval estimate would get narrower. 99% confidence interval estimate of the population mean has a margin of error of 1.55. If the confidence level decreased to 95%, the margin of error would get smaller and the width of the interval estimate would get narrower. 99% confidence interval estimate of the population mean has a margin of error of 1.55. If the confidence level decreased to 95%, the margin of error would get larger and the width of the interval estimate would get larger.

Solution

Note that              
              
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 =    0.005          
X = sample mean =    15.3          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    3.8          
n = sample size =    40          
              
Thus,              
              
Lower bound =    13.75235738          
Upper bound =    16.84764262          
              
Thus, the confidence interval is              
              
(   13.75235738   ,   16.84764262   )

Thus,

E = margin of error = (upper - lower)/2 = 1.547642618 = 1.55

Now, lower confidence levels yield NARROWER confidence intervals.

Thus, the answer is OPTION B. [ANSWER, B]