Tdistribution to construct a confidence interval question

joe98
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The following are summary statistics for a data set. Would it be appropriate to use the
Student's t- distribution to construct a confidence interval for these data?
Explain clearly!

Sample size = 10
Mean = 8.905
Median = 6.105
Standard deviation = 9.690
Minimum = 0.512
Maximum = 39.920
Q1 = 1.967
Q3 = 8.103

My answer would be it would be appropriate since n is less than 40(small sample size), we can use t distribution

also the population should be normally distributed

Any other explanations guys
 
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Any thoughts guys?
 
Any ideas guys?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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