Answer:
Explanation:
Given that a student conducted a study and reported that the 95% confidence interval for the mean ranged from 46 to 54
Hence width of interval = 8
Margin of error = ±1/2 (width)=±4
Margin of error = Z critical for 95%( std error)

Equate the two terms for margin of error we get
(\frac{16}{\sqrt{n}=

n=61.4656
So sample size is approximately 61.