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 54.2 minutes, with a standard deviation of 14.0 minutes. What is the 99% confidence interval (z-score = 2.58) 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 mc019-1.jpg.

49.1 minutes to 59.3 minutes

50.3 minutes to 58.1 minutes

54.2 minutes to 58.1 minutes

54.2 minutes to 59.3 minutes

Margin of error = z δ/√n

sample population = 50
mean = 54.2 minutes
standard deviation = 14.0 minutes
z-score = 2.58

2.58 (14/√50) = 2.58 14/7.07 = 2.58 1.98 = 5.1084 or 5.11

54.2 + 5.11 = 59.31 minutes
54.2 - 5.11 = 49.09 minutes

Answer is: 49.1 minutes to 59.3 minutes

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AFOKE88 AFOKE88 · Answer 2 · 2022-05-18T05:24:58-04:00

The confidence interval of the given statistics when calculated with the given parameters is; (49.1 minutes to 59.3 minutes)

How to find the confidence interval?

Formula for Margin of error is;

M = z * σ/√n

We are given;

Sample size; n = 50

mean; x' = 54.2 minutes

standard deviation; σ = 14.0 minutes

z-score = 2.58

Thus;

M = 2.58 × (14/√50)

M = 5.11

Confidence interval is;

CI = x' ± M

CI = (54.2 + 5.1), (54.2 - 5.1)

CI = (49.1 minutes, 59.3 minutes)

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