Question

Suppose that at a state college, a random sample of 41 students is drawn, and each of the 41 students in the sample is asked to measure the length of their right foot in centimeters. A 95% confidence interval for the mean foot length for students at this college turns out to be (21.709, 25.091). If instead a 90% confidence interval was calculated, how would it differ from the 95% confidence interval?

Answer 1

Answer:

The  confidence interval for 90% confidence would be narrower than the 95% confidence

Step-by-step explanation:

From the question we are told that

  The  sample size is n = 41

   

For a 95% confidence the level of significance is  $\alpha = [100 - 95]\% = 0.05$ and

the critical value  of  $\frac{\alpha }{2}$  is   $Z_{\frac{\alpha }{2} } =Z_{\frac{0.05 }{2} }= 1.96$

For a 90% confidence the level of significance is  $\alpha = [100 - 90]\% = 0.10$ and

the critical value  of  $\frac{\alpha }{2}$  is   $Z_{\frac{\alpha }{2} } =Z_{\frac{0.10 }{2} }= 1.645$

So we see with decreasing confidence level the critical value  decrease

Now the margin of error is mathematically represented as

         $E = Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{n} }$

given that other values are constant and only $Z_{\frac{\alpha }{2} }$ is varying we have that

         $E\ \ \alpha \ \ Z_{\frac{\alpha }{2} }$

Hence for  reducing confidence level the margin of error will be reducing

  The  confidence interval is mathematically represented as

        $\= x - E < \mu < \= x + E$

Now looking at the above formula and information that we have deduced so far we can infer that as the confidence level reduces , the critical value  reduces, the margin of error  reduces and the confidence interval becomes narrower