Mathematics: Explore parameter changes page 58

Explore parameter changes page 58 - id585028638, gungamer720

Ответ:

leighsa

08.09.2020

Mathematics

b) 95 percent confidence interval for this single-sample t test

[11.64, 16.36]

Step-by-step explanation:

Explanation:-

Given data a study of 62 college students finds that their average interest rate is 14 percent with a standard deviation of 9.3 percent.

Sample size 'n' =62

sample mean x⁻ = 14

sample standard deviation 'S' = 9.3

95 percent confidence interval for this single-sample t test

The values are $(x^{-} - t_{0.05} \frac{S}{\sqrt{n} } ,x^{-}+t_{0.05}\frac{S}{\sqrt{n} })$ the 95 percent confidence interval for the population mean 'μ'

Degrees of freedom γ=n-1=62-1=61

t₀.₀₅ = 1.9996 at 61 degrees of freedom

$(14 - 1.9996\frac{9.3}{\sqrt{62} } ,14+1.9996\frac{9.3}{\sqrt{62} })$

(14-2.361 , 14 + 2.361)

[(11.64 , 16.36]

Conclusion:-

95 percent confidence interval for this single-sample t test

[11.64, 16.36]

Assume that the national credit card interest rate is 12.83 percent. A study of 62 college students

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Explore parameter changes page 58

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