Question

3. The attendance at the George Brown Huskies basketball game last night was 400. A random sample of 50 of those in attendance revealed that the mean number of soft drinks consumer per person was 1.86 with a standard deviation of 0.59. Develop a 99 percent confidence interval for the mean number of soft drinks consumed per person.

Answer 1

From the given data sample data,

Sample size n = 50,

Sample mean X = 1.86
Sample standard deviation s = 0.59

Level of confidence = 99 percent. So 1 - α = 0.99, α =0.01 and α/2 = 0.005
Therefore, z_α/2 = z_0.005 = 2.58

Margin of error = MOE = E =z_α/2 s/√n = 2.58*0.59/√50= 0.215

99% Confidence Interval = (X ± E) = (1.86 ± 0.215) = (1.645, 2.075).

The 99% confidence interval for the mean number of soft drinks consumed per person is (1.645, 2.075).