Question

There is a sample of 85 people from a large city. From this sample, the mean number of glasses of water consumed per day is 3.65. Assume that the population is approximately normal and has a standard dev. of 2.5 glasses consumed per day.

Construct a 93% confidence interval for the mean number of glasses of water consumed among all residents of the city.

Also state what procedure you're using, the critical value, degrees of freedom (if applicable), and the 93% confidence interval's upper and lower bounds.

Thank you!

Answer 1

Standard deviation is known so we use z multiplier (critical value)

Form Z table,

Z$\alpha$/2 = 1..812

margin of error E = Z$\alpha$/2 * $\sigma$ / sqrt (n)

= 1.812 * 2.5 / sqrt(85)

= 0.491

93% confidence interval for $\mu$ is

$\bar{x}$ - E < $\mu$ < $\bar{x}$ + E

3.65 - 0.491 < $\mu$ < 3.65 + 0.491

3.159 < $\mu$ < 4.141

93% CI is ( 3.159 , 4.141 )

Lower bound = 3.159

Upper bound = 4.141