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dezoksy [38]
3 years ago
9

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 r

andomly 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.
Mathematics
2 answers:
alexdok [17]3 years ago
8 0

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)

Natalija [7]3 years ago
4 0

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].

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