Question

4. Spring break can be a very expensive holiday. A sample of 90 students is surveyed, and the average amount spent by students on travel and beverages is $601.00. The sample standard deviation is approximately $394.34. Construct a 95% confidence interval for the population mean amount of money spent by spring breakers. Draw graph.

Answer 1

Solution :

Given that,

= 601.00

s = 394.34

n = 90

Degrees of freedom = df = n - 1 = 90 - 1 = 89

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,89 =2.797  

Margin of error = E = t/2,df * (s /n)

= 2.797 * (394.34 / 90)

=85.59

Margin of error = 85.59

The 95% confidence interval estimate of the population mean is,

- E < < + E

601.00 - 85.59 < < 601.00 + 85.59

518.41 < < 683.59

(518.41, 683.59 )