A simple random sample of 50 adults was surveyed, and it was found that the mean amount of time that they spend surfing the Internet per day is 54 point 2 minutes, with a standard deviation of 14 point 0 minutes. What is the 99 percent confidence interval open paren z - score is equal to 2 point 5 8 close paren for the number of minutes that an adult spends surfing the Internet per day? Remember, the margin of error, ME , can be determined using the formula ME is equal to the fraction with numerator z times s and denominator square root of n
Answer
A
49 point 1 minutes to 59 point 3 minutes
B
54 point 2 minutes to 59 point 3 minutes
C
54 point 2 minutes to 58 point 1 minutes
D
50 point 3 minutes to 58 point 1 minutes
Explanation
Use the margin-of-error formula: ME = 2.58 × 14.0 ÷ √50 ≈ 5.1 minutes. Add and subtract this from the sample mean of 54.2 minutes to get 54.2 − 5.1 ≈ 49.1 and 54.2 + 5.1 ≈ 59.3. The 99 percent confidence interval is 49.1 to 59.3 minutes, choice A.
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