Question

You work for a consumer advocate agency and want to find the mean repair cost of a washing machine. As part of your study, you randomly select 30 repair costs and find the mean to be $100.00. The standard deviation of the sample is $25.20. Calculate a 90% confidence interval for the population mean.

Answer 1

Answer:

90% confidence interval for the population mean is between a lower limit of $92.18 and an upper limit of $107.82.

Step-by-step explanation:

Confidence interval for a population mean is given as mean +/- margin of error (E)

mean = $100

sd = $25.20

n = 30

degree of freedom = n-1 = 30-1 = 29

confidence level (C) = 90% = 0.9

significance level = 1 - C = 1 - 0.9 = 0.1 = 10%

critical value (t) corresponding to 29 degrees of freedom and 10% significance level is 1.699

E = t×sd/√n = 1.699×25.20/√30 = $7.82

Lower limit of population mean = mean - E = 100 - 7.82 = $92.18

Upper limit of population mean = mean + E = 100 + 7.82 = $107.82

90% confidence interval is ($92.18, $107.82)

Answer 2

Answer:

90% confidence interval for the population mean is [92.18 , 107.82].

Step-by-step explanation:

We are given that as part of your study, you randomly select 30 repair costs and find the mean to be $100.00. The standard deviation of the sample is $25.20.

So, the pivotal quantity for 90% confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean = $100

            s = sample standard deviation = $25.20

            n = sample size = 30

            $\mu$ = population mean

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.699 < $t_2_9$ < 1.699) = 0.90

P(-1.699 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 1.699) = 0.90

P( $-1.699 \times {\frac{s}{\sqrt{n} }$ < ${\bar X - \mu}$ < $1.699 \times {\frac{s}{\sqrt{n} }$ ) = 0.90

P( $\bar X - 1.699 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X + 1.699 \times {\frac{s}{\sqrt{n} }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X - 1.699 \times {\frac{s}{\sqrt{n} }$ , $\bar X + 1.699 \times {\frac{s}{\sqrt{n} }$ ]

                                                = [ $100 - 1.699 \times {\frac{25.20}{\sqrt{30} }$ , $100 + 1.699 \times {\frac{25.20}{\sqrt{30} }$ ]

                                                = [92.18 , 107.82]

Therefore, 90% confidence interval for the population mean is [92.18 , 107.82].