Question

In a sample of 60 electric motors, the average efficiency (in percent) was 85 and the standard deviation was 2. Section 05.01 Exercise 12.a - Compute confidence interval; Find necessary sample size Find a 95% confidence interval for the mean efficiency. Round the answers to three decimal places.

Answer 1

Answer:

95% confidence interval for the mean efficiency is [84.483 , 85.517].

Step-by-step explanation:

We are given that in a sample of 60 electric motors, the average efficiency (in percent) was 85 and the standard deviation was 2.

So, the pivotal quantity for 95% confidence interval for the population mean efficiency is given by;

          P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\mu$ = sample average efficiency = 85

            $\sigma$ = sample standard deviation = 2

            n = sample of motors = 60

            $\mu$ = population mean efficiency

So, 95% confidence interval for the mean efficiency, $\mu$ is ;

P(-2.0009 < $t_5_9$ < 2.0009) = 0.95

P(-2.0009  < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.0009 ) = 0.95

P( $-2.0009 \times {\frac{s}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.0009 \times {\frac{s}{\sqrt{n} }$ ) = 0.95

P( $\bar X -2.0009 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.0009 \times {\frac{s}{\sqrt{n} }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X -2.0009 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.0009 \times {\frac{s}{\sqrt{n} }$ ]

                                                 = [ $85 -2.0009 \times {\frac{2}{\sqrt{60} }$ , $85 +2.0009 \times {\frac{2}{\sqrt{60} }$ ]

                                                 = [84.483 , 85.517]

Therefore, 95% confidence interval for the population mean efficiency is [84.483 , 85.517].