Answer to Question #223434 in Statistics and Probability for Ruben

Question

(a)   A survey of 25 individuals who passed the ACCA exam showed that average annual salary is RM150,000 and standard deviation RM15,000.

(i)                    Find the degrees of freedom.

 

(ii)                  Construct 99% confidence interval for the true average salary.

Answer 1

(i) $df=n-1=25-1=24$ degrees of freedom.

(ii) The critical value for $\alpha=0.01$ and $df=24$ degrees of freedom is $t_c=2.79694.$

The corresponding confidence interval is computed as shown below:

$$CI=(x-t_c\times\dfrac{s}{\sqrt{n}},x+t_c\times\dfrac{s}{\sqrt{n}} )$$

$$=(150000-2.79694\times\dfrac{15000}{\sqrt{25}},$$

$$150000+2.79694\times\dfrac{15000}{\sqrt{25}})$$

$$=(141609.18, 158390.82)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $141609.18<\mu<158390.82,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(141609.18, 158390.82).$