Question

Suppose a 95% confidence interval for obtained from a random sample of size 13 is (3.5990, 19.0736). Find the lower bound of a 90% confidence interval for (round off to the nearest integer).

Answer 1

Answer:

The lower bound of a 90% confidence interval = 4.83

Step-by-step explanation:

Let us assume that the 95% confidence interval for population mean is constructed from a random sample of size 13 which is (3.5990, 19.0736).

We calculate 95% confidence interval for population mean by;

              x bar $\pm$ 1.96$\frac{\sigma}{\sqrt{n} }$ , where xbar = sample mean or observed mean

                                                      $\sigma$      = Population standard deviation

                                                      n      = sample size

                                                    1.96 = It represent that at 2.5% level of  

                                                    significance the area of z score will be 1.96.

So (3.5990, 19.0736) = x bar $\pm$ 1.96$\frac{\sigma}{\sqrt{13} }$ , which further represent

       x bar - 1.96$\frac{\sigma}{\sqrt{13} }$  =  3.5990     Equation 1

       x bar + 1.96$\frac{\sigma}{\sqrt{13} }$  =  19.0736     Equation 2

Solving these two above questions we get x bar = 11.336   and $\sigma$ = 14.233

Now Similarly the 90% Confidence Interval = x bar $\pm$ 1.6449$\frac{\sigma}{\sqrt{n} }$

So, the lower bound for this confidence interval is = x bar - 1.6449$\frac{\sigma}{\sqrt{13} }$

                                                                                   = 11.336 - 1.6449$\frac{14.233}{\sqrt{13} }$

                                                                                   = 4.843 or 4.83

Therefore, the lower bound of a 90% confidence interval is 4.83.