Question

A random sample of 144 observations has a mean of 20, a median of 21, and a mode of 22. The population standard deviation is known to equal 4.8. The 95.44% confidence interval for the population mean is _____.

Answer 1

Answer:

95.44% confidence interval for the population mean is [19.2 , 20.8].

Step-by-step explanation:

We are given that a random sample of 144 observations has a mean of 20, a median of 21, and a mode of 22. The population standard deviation is known to equal 4.8.

Also, assuming that the data follows normal distribution.

So, the pivotal quantity for 95.44% confidence interval for the population mean is given by;

            P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean = 20

            $\sigma$ = population standard deviation = 4.8

            n = sample of obs. = 144

            $\mu$ = population mean

So, 95.44% confidence interval for the population mean, $\mu$ is ;

P(-1.9991 < N(0,1) < 1.9991) = 0.9544

P(-1.9991 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.9991) = 0.9544

P( $-1.9991 \times {\frac{4.8}{\sqrt{144} }$ < ${\bar X - \mu}$ < $1.9991 \times {\frac{4.8}{\sqrt{144} }$ ) = 0.9544

P( $\bar X - 1.9991 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X + 1.9991 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.9544

95.44% confidence interval for $\mu$ = [ $\bar X - 1.9991 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X + 1.9991 \times {\frac{\sigma}{\sqrt{n} }$ ]

                                                       = [ $20 - 1.9991 \times {\frac{4.8}{\sqrt{144} }$ , $20 + 1.9991 \times {\frac{4.8}{\sqrt{144} }$ ]

                                                       = [19.2 , 20.8]

Therefore, 95.44% confidence interval for the population mean is [19.2 , 20.8].