Answer:
The confidence interval is [tex]25.16 < \mu < 26.85[/tex]
Step-by-step explanation:
From the question we are given a data set
25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.
The mean of the this sample data is
[tex]\= x = \frac{\sum x_i}{n}[/tex]
where is the sample size with values n = 10
[tex]\= x = \frac{25.5+ 26.1+ 26.8+23.2+ 24.2+ 28.4+ 25.0+ 27.8+ 27.3+ 25.7}{10}[/tex]
[tex]\= x = 26[/tex]
The standard deviation is evaluated as
[tex]\sigma = \sqrt{\frac{\sum (x-\= x)}{n} }[/tex]
substituting values
[tex]= \sqrt{\frac{ ( 25.5-26)^2, (26.1-26)^2, (26.8-26)^2, (23.2-26)^2}{10} }[/tex]
[tex]\cdot \ \cdot \ \cdot \sqrt{\frac{ ( 24.2-26)^2, (28.4-26)^2+( 25.0-26)^2+ (27.8-26)^2+( 27.3-26)^2+( 25.7-26)^2}{10} }[/tex]
[tex]\sigma = 1.625[/tex]
The confidence level is given as 90% hence the level of significance is calculated as
[tex]\alpha = 100 -90[/tex]
[tex]\alpha =10[/tex]%
[tex]\alpha = 0.10[/tex]
Now the critical values of [tex]\frac{\alpha }{2}[/tex] is obtained from the normal distribution table as
[tex]Z_{\frac{\alpha }{2} } = 1.645[/tex]
The reason we are obtaining the critical values of [tex]\frac{\alpha }{2}[/tex] instead of [tex]\alpha[/tex] is because we are considering two tails of the area under the normal curve
The margin of error is evaluated as
[tex]MOE = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
substituting values
[tex]MOE = 1.645 * \frac{1.625 }{\sqrt{10} }[/tex]
[tex]MOE = 0.845[/tex]
The 90%, two sided confidence interval is mathematically evaluated as
[tex]\= x - MOE < \mu < \= x + MOE[/tex]
[tex]26 - 0.845 < \mu < 26 + 0.845[/tex]
[tex]25.16 < \mu < 26.85[/tex]
Given that the lower and the upper limit is greater than 25 then we can assure the manufactures that the battery life exceeds 25 hours
