Question

19. FHSU wants to know if their students tend to drink more coffee than the national average. They ask a random sample of 75 students how many cups of coffee they drink each day and find the average to be 3.8 cups with a standard deviation of 1.5 20. Redo #19, only this time assume the value 1.5 is the population standard deviation. Then, state this interval below within an interpretive sentence tied to the given context. Please explain 20

Answer 1

Since the confidence level is not stated in the problem, I am assuming it to be 95%

(19)

n = 75     

x-bar = 3.8     

s = 1.5     

% = 95     

Standard Error, SE = s/√n =    1.5/√75 = 0.173205081

Degrees of freedom = n - 1 =   75 -1 = 74   

t- score = 1.992543466     

Width of the confidence interval = t * SE =     1.99254346583172 * 0.173205080756888 = 0.345118652

Lower Limit of the confidence interval = x-bar - width =      3.8 - 0.345118651910993 = 3.454881348

Upper Limit of the confidence interval = x-bar + width =      3.8 + 0.345118651910993 = 4.145118652

The 95% confidence interval is [3.45, 4.15]

We are 95% confident that the true average number of coffees consumed lies in the above interval

(20)

n = 75     

x-bar = 3.8     

s = 1.5     

% = 95     

Standard Error, SE = σ/√n =    1.5 /√75 = 0.173205081

z- score = 1.959963985     

Width of the confidence interval = z * SE =     1.95996398454005 * 0.173205080756888 = 0.33947572

Lower Limit of the confidence interval = x-bar - width =      3.8 - 0.339475720222852 = 3.46052428

Upper Limit of the confidence interval = x-bar + width =      3.8 + 0.339475720222852 = 4.13947572

The 95% confidence interval is [3.46, 4.14]

We are 95% confident that the true average number of coffees consumed lies in the above interval.