Question

1. A University of Arkansas student running for president of her sorority wanted to identify the issues are most important to the members of her sorority. To do so, she created a survey that asks various questions concerning the sorority. Thirty-seven out of 120 members completed the survey.   

   The first question on the survey asked the members to rate the quality of food provided in the sorority house. The answers can range from 1 (very poor) to 5 (very good). The average rating was 3.7 with a standard deviation of 0.5. Construct of 90% confidence interval for the rating and identify which value (i.e., t or z) you used when constructing it.

   		[3.62, 3.78] and used a z-value
   		[3.56, 3.84] and used a z-value
   		[3.56, 3.84] and used a t-value
   		[3.62, 3.78] and used a t-value

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 3.7

sample standard deviation = s = 0.5

sample size = n = 37

Degrees of freedom = df = n - 1 = 37 - 1 = 36

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t0.05,36 = 1.688

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

= 1.688 * ( 0.5 / 37)

Margin of error = E = 0.14

The 90% confidence interval estimate of the population mean is,

  ± E  

= 3.7   ± 0.14

= ( 3.56, 3.84 )

[3.56, 3.84] and used a t-value