Question

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 $me = \frac{z\cdot s}{\sqrt{n}}$. 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

Explanation:

Step1: Identify given values

$z = 2.58$, $s=14.0$, $n = 50$, $\bar{x}=54.2$

Step2: Calculate the margin of error

$ME=\frac{z\cdot s}{\sqrt{n}}=\frac{2.58\times14.0}{\sqrt{50}}\approx\frac{36.12}{7.07}\approx5.11$

Step3: Calculate the lower - bound of the confidence interval

Lower bound $=\bar{x}-ME=54.2 - 5.11=49.09\approx49.1$

Step4: Calculate the upper - bound of the confidence interval

Upper bound $=\bar{x}+ME=54.2+5.11 = 59.31\approx59.3$

Answer:

49.1 minutes to 59.3 minutes