Respuesta :
Given Information:
Sample size = n = 25
Mean age of client = 46 years
Standard deviation of age of client = 5 years
Confidence level = 95%
Required Information:
Width of the confidence interval = ?
Answer:
[tex]$ \text {width of CI } = \pm 2.064 $\\\\[/tex]
Step-by-step explanation:
The width for the true mean client age is given by
[tex]$ \text {width of CI } =\pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $[/tex]
Where n is the sample size, s is the standard deviation of age of client, and [tex]t_{\alpha/2}[/tex] is the t-score corresponding to 95% confidence level.
The t-score corresponding to 95% confidence level is
Significance level = 1 - 0.95 = 0.05/2 = 0.025
Degree of freedom = n - 1 = 25 - 1 = 24
From the t-table at α = 0.025 and DF = 24
t-score = 2.064
[tex]$ \text {width of CI } = \pm (2.064)(\frac{5}{\sqrt{25} } ) $\\\\[/tex]
[tex]$ \text {width of CI } =\pm (2.064)(\frac{5}{5 } ) $\\\\[/tex]
[tex]$ \text {width of CI } = \pm (2.064)(1) $\\\\[/tex]
[tex]$ \text {width of CI } = \pm 2.064 $\\\\[/tex]
Therefore, width of the 95% confidence Interval for the true mean client age is approximately ±2.064.
Bonus:
The corresponding 95% confidence interval is given by
[tex]$ \text {Confidence Interval } = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $[/tex]
Where x_bar is the mean client age
[tex]$ \text {Confidence Interval } = 46 \pm 2.064 $[/tex]
[tex]$ \text {Confidence Interval } = (43.936, 48.064) $[/tex]
[tex]$ \text {Confidence Interval } = (44, 48) $[/tex]
Which means that we are 95% sure that the true mean client age is within the range of (44, 48) years.
