Answer :
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
[tex]x^{-} ± t_{\alpha } \frac{S}{\sqrt{n} }[/tex]
the degrees of freedom γ=n-1 =20-1=19
t-table 2.093
[tex](x^{-} - t_{\alpha } \frac{S}{\sqrt{n} },x^{-} + t_{\alpha } \frac{S}{\sqrt{n} })[/tex]
[tex](2.92 - 2.093(\frac{0.16}{\sqrt{20} } ,2.92+2.093(\frac{0.16}{\sqrt{20} } )[/tex]
(2.92-0.0748,2.92+0.0748)
(2.84 ,2.99)
Therefore the 95% of confidence intervals
(2.84 ,2.99)
