You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of business​ days, the mean closing price of a certain stock was ​$. Assume the population standard deviation is ​$. The​ 90% confidence interval is ​( nothing​, nothing​). ​(Round to two decimal places as​ needed.) The​ 95% confidence interval is ​( nothing​, nothing​). ​(Round to two decimal places as​ needed.) Which interval is​ wider? Choose the correct answer below

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Complete Question

The  complete question is shown on the first uploaded image

Answer:

The 90% confidence interval is  [tex][108.165 ,112.895][/tex]

The  95%  confidence interval is [tex][107.7123 ,113.3477][/tex]

The  correct option is  D

Step-by-step explanation:

From the question we are told that

    The sample size is  n =  48

     The sample  mean is  [tex]\= x = \$ 110.53[/tex]

    The standard deviation is  [tex]\sigma = \$ 9.96[/tex]

Considering first question

  Given that the confidence level is  90% then the level of significance is mathematically represented as

            [tex]\alpha = (100 - 90)\%[/tex]

           [tex]\alpha = 0.10[/tex]

The  critical value  of  [tex]\frac{\alpha }{2}[/tex] from the  normal distribution table is  

          [tex]Z_{\frac{\alpha }{2} } = 1.645[/tex]

Generally the margin of error is mathematically represented as

            [tex]E = ZZ_{ \frac{x}{y} } * \frac{\sigma}{ \sqrt{n} }[/tex]

             [tex]E = 1.645 * \frac{9.96}{ \sqrt{ 48} }[/tex]

             [tex]E = 2.365[/tex]

The  90% confidence interval is  

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

=>    [tex]110.53 - 2.365 < \mu < 110.53 + 2.365[/tex]

=>    [tex]108.165 < \mu < 112.895[/tex]

Considering second question

  Given that the confidence level is  95% then the level of significance is mathematically represented as

            [tex]\alpha = (100 - 95)\%[/tex]

           [tex]\alpha = 0.05[/tex]

The  critical value  of  [tex]\frac{\alpha }{2}[/tex] from the  normal distribution table is  

          [tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]

Generally the margin of error is mathematically represented as

            [tex]E = Z_{ \frac{x}{y} } * \frac{\sigma}{ \sqrt{n} }[/tex]

             [tex]E = 1.96 * \frac{9.96}{ \sqrt{ 48} }[/tex]

             [tex]E = 2.8177[/tex]

The  95% confidence interval is  

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

=>    [tex]110.53 - 2.8177 < \mu < 110.53 + 2.8177[/tex]

=>    [tex]107.7123 < \mu < 113.3477[/tex]

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