Question

The university would like to conduct a study to estimate the true proportion of all university students who have student loans. According to the study, in a random sample of 215 university students, 86 have student loans.

(a) Construct a 99% confidence interval for estimating the true proportion of all university students who have student loans (2 marks)

(b) Provide an interpretation of the confidence interval in part (a). (1mark)

(c) Conduct an appropriate hypothesis test, at the 1% level of significance to test the claim that more than 30% of all university students have student loans.

1. Provide the hypothesis statement

2. Calculate the test statistic value

3. Determine the probability value

Note: if you need to use symbols , please use

• "u" for population mean "μ",

• Ho and Ha for  for the null and alternate hypothesis,

• "Y-hat" for "ŷ", "alpha" for α

Please provide your answers to the above questions by typing your answers using simple text. You need not show the work in detail.

Answer 1

a)

Level of Significance,   α =    0.01          
Number of Items of Interest,   x =   86          
Sample Size,   n =    215          
                  
Sample Proportion ,    p̂ = x/n =    0.4000          
z -value =   Zα/2 =    2.576   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.03341          
margin of error , E = Z*SE =    2.576   *   0.03341   =   0.0861
                  
99%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.40000   -   0.08606   =   0.31394
Interval Upper Limit = p̂ + E =   0.40000   +   0.08606   =   0.48606
                  
99%   confidence interval is (   0.314   < p <    0.486   )

b) there is 95% confidence that true porportion lies within confidence interval

c)

Ho :   p =    0.3                  
H1 :   p >   0.3       (Right tail test)          
                          
Level of Significance,   α =    0.01                  
Number of Items of Interest,   x =   86                  
Sample Size,   n =    215                  
                          
Sample Proportion ,    p̂ = x/n =    0.4000                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0313                  
Z Test Statistic = ( p̂-p)/SE = (   0.4000   -   0.3   ) /   0.0313   =   3.1997
                          

                          
p-Value   =   0.0007   [Excel function =NORMSDIST(-z)              
Decision:   p-value<α , reject null hypothesis                       
There is enough evidence to support the claim