Question

The GPA of accounting students in a university is known to be normally distributed. A random sample of 20 accounting students results in a mean of 2.92 and a standard deviation of 0.16. Construct the 95% confidence interval for the mean GPA of all accounting students at this university.

Answer 1

Answer:

The 95% of confidence intervals

(2.84 ,2.99)

Step-by-step explanation:

A random sample of 20 accounting students results in a mean of 2.92 and a standard deviation of 0.16

given small sample size n =20

sample mean x⁻ =2.92

sample standard deviation 'S' =0.16

level of significance ∝ =  0.95

The 95% of confidence intervals

$x^{-} ± t_{\alpha } \frac{S}{\sqrt{n} }$

the degrees of freedom γ=n-1 =20-1=19

t-table 2.093

$(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} },x^{-} + t_{\alpha } \frac{S}{\sqrt{n} })$

$(2.92 - 2.093(\frac{0.16}{\sqrt{20} } ,2.92+2.093(\frac{0.16}{\sqrt{20} } )$

(2.92-0.0748,2.92+0.0748)

(2.84 ,2.99)

Therefore the 95% of confidence intervals

(2.84 ,2.99)