Thirty randomly selected students took the calculus final. if the sample mean was 95 and the standard deviation was 6.6, construct a 99% confidence interval of the mean score of all students. assume that the population has a normal distribution

Given:
n = 30 students
sample mean (X) = 95
standard deviation (s) = 6.6

We use the t distribution.

99% interval means 0.99 = 1 - α ; α = 0.01
tα/2 = t0.005 = invT(0.995,29) = 2.75638

X - t a/2 * (s/√n) < μ < X + t a/2 * (s/√n)

95 - 2.75638*(6.6/√30) < μ < 95 + 2.75638*(6.6/√30)
95 - 3 < μ < 95 + 3

92 < μ < 98

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nishantwork777 nishantwork777 · Answer 2 · 2021-10-26T07:38:51+02:00

By assuming that, the population has a normal distribution, the value of [tex]\mu[/tex] can be calculated for the standard deviation of 6.6, The random distribution will have [tex]\mu[/tex], as,

[tex]92<\mu<95[/tex]

Given information:

Sample mean [tex]=95[/tex]

The standard deviation [tex]=6.6[/tex]

Now using the t-Distribution

99% of interval refers 0.99

[tex]0.99=1-\alpha\\\alpha=0.01[/tex]

[tex]t(\alpha/2)=t'(0.995)=2.756[/tex]

Now using the equation:

[tex]X-t(\alpha/2)\times (s/\sqrt{n})<\mu<X+t(\alpha/2)\times (s/\sqrt{n})[/tex]

[tex]95-2.7563\times(6.6/\sqrt{30} )<\mu<95+2.7563\times(6.6/\sqrt{30})[/tex]

[tex]95-3<\mu<95+3[/tex]

[tex]92<\mu<98[/tex]

Hence,

The random distribution will have [tex]\mu[/tex]

[tex]92<\mu<95[/tex]

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