Answer:
The correct option is B
Step-by-step explanation:
The formula for confidence interval for the mean is
\(Interval=\mu\pm z*\times \frac{\sigma}{\sqrt{n}}\)
Where, μ is population mean, σ is standard deviation, n is sample size and z* is z-score at given confidence interval.
From the z-table the value of z-score at 98% confidence interval is 2.33.
From the given information it is clear that
\(\mu=50\)
\(\sigma=6\)
\(n=100\)
The 98% confidence interval for the mean hours devoted to social networking in January is
\(Interval=50\pm 2.33\times \frac{6}{\sqrt{100}}\)
\(Interval=50\pm 2.33\times 0.6\)
\(Interval=50\pm 1.398\)
\(Interval=[50-1.398,50+1.398]\)
\(Interval=[48.602,51.398]\)
\(Interval=[48.60,51.40]\)
Therefore the 98% confidence interval ranges from 48.60 to 51.40 hours. Option B is correct.
