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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?

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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  [tex]\alpha = [100 - 95]\% = 0.05[/tex] and

the critical value  of  [tex]\frac{\alpha }{2}[/tex]  is   [tex]Z_{\frac{\alpha }{2} } =Z_{\frac{0.05 }{2} }= 1.96[/tex]

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

the critical value  of  [tex]\frac{\alpha }{2}[/tex]  is   [tex]Z_{\frac{\alpha }{2} } =Z_{\frac{0.10 }{2} }= 1.645[/tex]

So we see with decreasing confidence level the critical value  decrease

Now the margin of error is mathematically represented as

         [tex]E = Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{n} }[/tex]

given that other values are constant and only [tex]Z_{\frac{\alpha }{2} }[/tex] is varying we have that

         [tex]E\ \ \alpha \ \ Z_{\frac{\alpha }{2} }[/tex]

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

  The  confidence interval is mathematically represented as

        [tex]\= x - E < \mu < \= x + E[/tex]

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

Using confidence interval concepts, according to the margin of error, the 90% confidence interval would be narrower than the 95% confidence interval.

The margin of error of a confidence interval is given by:

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which:

  • z is the critical value.
  • [tex]\sigma[/tex] is the population standard deviation.
  • n is the sample size.

The higher the margin of error, the wider the interval is.

  • The lower the confidence interval, the lower the critical value is. Hence, a 90% confidence interval has a lower critical value than a 95% confidence interval, leading to a smaller margin of error and a narrower interval.

A similar problem is given at https://brainly.com/question/22966154

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