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

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.  =    ~  

where, = sample proportion of defective items = 12,898

             s = sample standard deviation = 7,719

            n = sample size = 5

              = 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, is ;

P(-2.776 < < 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 < < 2.776) = 0.95

P( < < ) = 0.95

P( < < ) = 0.95

95% confidence interval for = [ , ]

                                = [ , ]

                               = [3315.13, 22480.87]

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

45

step-by-step explanation:

hi there!

the answer will be d: 2