Data on investments in the​ high-tech industry by venture capitalists are compiled by a corporation. A random sample of 18​venture-capital investments in a certain business sector yielded the accompanying​ data, in millions of dollars. Determine and interpret a 95​%confidence interval for the mean​ amount, mu​,of all​ venture-capital investments in this business sector. Assume that the population standard deviation is ​$1.70million.​ (Note: The sum of the data is ​$102.52​million.)

Answer:

$4.911  million or  $6.481 million

Thus, we are 95% confident that the mean amount of all venture-capital investments in the high-tech industry is somewhere between $4.911 million and $6.481 million.

Step-by-step explanation:

Given that:

sample size n = 18

standard deviation σ = 1.70

confidence interval = 95%

Sample mean [tex]\overline x =\dfrac{ \sum x }{n}[/tex]

[tex]\overline x =\dfrac{ 102.52 }{18}[/tex]

[tex]\overline x =[/tex] 5.696

The level of significance = 1 - C.I

= 1 - 0.95

= 0.05

The critical value of [tex]Z_{\alpha/2} = Z_{0.025} = 1.960[/tex] from the Z tables

The 95% C.I for the mean is;

[tex]= \overline x \pm Z_{\alpha/2} \times \dfrac{\sigma}{\sqrt{n}}[/tex]

[tex]=5.696 \pm 1.960 \times \dfrac{1.70}{\sqrt{18}}[/tex]

[tex]=5.696 \pm 1.960 \times \dfrac{1.70}{4.243}[/tex]

[tex]=5.696 \pm 1.960 \times 0.4007[/tex]

= 5.696 ± 0.785372

= (5.696 - 0.785372 , 5.696 + 0.785372 )

= ( 4.910628 , 6.481372  )

≅ $4.911  million or  $6.481 million.

Thus, we are 95% confident that the mean amount of all venture-capital investments in the high-tech industry is somewhere between $4.911 million and $6.481 million.

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