Question

A simple random sample of 144 items resulted in a sample mean of 1250.25 and a standard deviation of 480. Which of the following options gives the correct formula for a 99% confidence interval for the population mean

Answer 1

Answer:

Correct formula for a 99% confidence interval for the population mean = $[xbar - 2.61138*\frac{480}{\sqrt{144} } , xbar + 2.61138*\frac{480}{\sqrt{144} }]$

Step-by-step explanation:

We are provided a simple random sample of 144 items of which sample mean, xbar = 1250.25 and Standard Deviation, s = 480.

We know that  $\frac{xbar - \mu}{\frac{s}{\sqrt{n} } }$  follows $t_n_-_1$ i.e.;

                         $\frac{1250.25 - \mu}{\frac{480}{\sqrt{144} } }$ follows $t_1_4_3$

So, 99% confidence interval is given by ;

 P(-2.61138 < $t_1_4_3$ < 2.61138) = 0.99 {because at 143 degree of freedom t table

                                                       gives critical value of 2.61138 at 0.5% level}

Now, after further solving this we get,

99% confidence interval for the population mean =

             $[xbar - 2.61138*\frac{480}{\sqrt{144} } , xbar + 2.61138*\frac{480}{\sqrt{144} }]$

Since, in the question options are not given so this must be the correct formula  for calculating 99% confidence interval for the population mean.