Private colleges and universities rely on money contributed by individuals and corporations for their operating expenses. Much of this money is put into a fund called an​ endowment, and the college spends only the interest earned by the fund. A recent survey of 8 private colleges in the United States revealed the following endowments​ (in millions of​ dollars): 60.2,​ 47.0, 235.1,​ 490.0, 122.6,​ 177.5, 95.4, and 220.0. Summary statistics yield 180.975 and s143.042. Calculate a​ 95% confidence interval for the mean endowment of all the private colleges in the United States assuming a normal distribution for the endowments. Round to three decimal places.

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Answer:

The 95% confidence interval   [tex] 81.852  <  \mu <  280.098 [/tex]

Step-by-step explanation:

From the question we are told that

   The  sample mean is  [tex]\= x  =  180.975[/tex]

   The standard deviation is  [tex]\sigma  =  143.042[/tex]

   The sample size is n = 8

Given that the confidence level is  95%  then the level of significance is  

    [tex]\alpha =( 100- 95) \%[/tex]

=>  [tex]\alpha =0.05[/tex]    

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

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

Generally the margin of error is mathematically represented as  

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

=>  [tex]E = 1.96 *  \frac{ 143.042 }{\sqrt{8} }[/tex]

=>  [tex]E = 99.123 [/tex]

Generally 95% confidence interval is mathematically represented as  

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

=>    [tex]180.975 -99.123 <  p <  180.975 +99.123[/tex]

=>    [tex] 81.852  <  \mu <  280.098 [/tex]

   

Using the t-distribution, it is found that the 95% confidence interval for the mean endowment of all the private colleges in the United States assuming a normal distribution for the endowments, in millions of dollars, is (61.39, 300.56).

The information given is:

  • Sample mean of [tex]\overline{x} = 180.975[/tex].
  • Sample standard deviation of [tex]s = 143.042[/tex]
  • Sample size of [tex]n = 8[/tex].

We have the standard deviation for the sample, hence, the t-distribution is used.

The interval is given by:

[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]

The critical value for a two-tailed 95% confidence interval with 8 - 1 = 7 df is t = 2.3646.

Then:

[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 180.975 - 2.3646\frac{143.042}{\sqrt{8}} = 61.39[/tex]

[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 180.975 + 2.3646\frac{143.042}{\sqrt{8}} = 300.56[/tex]

The 95% confidence interval for the mean endowment of all the private colleges in the United States assuming a normal distribution for the endowments, in millions of dollars, is (61.39, 300.56).

To learn more about the t-distribution, you can take a look at https://brainly.com/question/25077184