Answer:
The 95% confidence interval for the mean is approximately;
(9.063, 9.737)
Step-by-step explanation:
The given parameters are;
The mean of the sample, [tex]\overline x[/tex] = 9.4
The sample variance, s² = 0.49
The sample size, n = 19
The confidence level = 95%
Therefore, we get;
[tex]CI=\bar{x}\pm t_{\alpha/2} \dfrac{s}{\sqrt{n}}[/tex]
The degrees of freedom, df = n - 1
∴ df = 19 - 1 = 18
From the t-table, the critical-t = 2.101
Therefore, we have;
[tex]CI=9.4\pm 2.101 \times \dfrac{\sqrt{0.49} }{\sqrt{19}}[/tex]
Therefore, we have;
9.4 - 2.101×√(0.49/19) < μ < 9.4 + 2.101×√(0.49/19)
9.0626 < μ < 9.737
∴ The 95% confidence interval for the mean ≈ (9.063, 9.737)
