Question

for the data values below construct a 95 confidence interval if the sample mean is known to be 12898 and the standard deviation is 7719

Answer 1

Answer:

A 95% confidence interval for the population mean is [3315.13, 22480.87] .

Step-by-step explanation:

We are given that for quality control purposes, we collect a sample of 200 items and find 24 defective items.

Firstly, the pivotal quantity for finding the 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 proportion of defective items = 12,898

             s = sample standard deviation = 7,719

            n = sample size = 5

             $\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics as we don't know about population standard deviation.

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

P(-2.776 < $t_4$ < 2.776) = 0.95  {As the critical value of t at 4 degrees of

                                               freedom are -2.776 & 2.776 with P = 2.5%}  

P(-2.776 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.776) = 0.95

P( $-2.776 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.776 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.776 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.776 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.776 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.776 \times {\frac{s}{\sqrt{n} } }$ ]

                                = [ $12,898-2.776 \times {\frac{7,719}{\sqrt{5} } }$ , $12,898+2.776 \times {\frac{7,719}{\sqrt{5} } }$ ]

                               = [3315.13, 22480.87]

Therefore, a 95% confidence interval for the population mean is [3315.13, 22480.87] .