"Suppose David owns a chain of 24 hr gyms. On average, members go to the gym 3.16 days a week with a standard deviation of 0.48 days. David decided to change the type of classes offered and wanted to see if the average member attendance would change. He randomly selected 250 members and found they attend the gym on average 3.13 days. Construct a two-tailed 90% confidence interval for the mean. Round your answer to two decimal places"

90% confidence interval for the mean is between a lower limit of 3.08 days and an upper limit of 3.18 days.

Step-by-step explanation:

Confidence interval is given as mean +/- margin of error (E)

mean = 3.13 days

sd = 0.48 day

n = 250

degree of freedom = n-1 = 250-1 = 249

confidence level (C) = 90% = 0.9

significance level = 1 - C = 1 - 0.9 = 0.1 = 10%

critical value (t) corresponding to 249 degrees of freedom and 10% significance level is 1.65153

E = t×sd/√n = 1.65153×0.48/√250 = 0.05 day

Lower limit of mean = mean - E = 3.13 - 0.05 = 3.08 days

Upper limit of mean = mean + E = 3.13 + 0.05 = 3.18 days

90% confidence interval is (3.08, 3.18)

he sells x mount because thats eqivilqant to the maz and then divided into quarters

step-by-step explanation: