Find a 95% confidence interval for population mean

chwala
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Homework Statement
see attached
Relevant Equations
t distribution
I am refreshing on this...

1662157094596.png


I think there is a mistake on the circled part in red...right? not correct symbol for sample mean...This is the part that i need clarity on.

1662157179394.png


The other steps to solution are pretty easy to follow...as long as one knows the t-formula and also the knowledge to interpret the t-distribution table with dof_{1} = ##9## and significance level i.e dof_{2}= ##0.025## that gives us the desired ##2.262##.

cheers
 
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The calculation is fine. The only thing I see wrong is that they wrote x for the sample mean instead of ##\bar x##. Possibly a typo or maybe the author is unable to write this symbol.
 
Mark44 said:
The calculation is fine. The only thing I see wrong is that they wrote x for the sample mean instead of ##\bar x##. Possibly a typo or maybe the author is unable to write this symbol.
@Mark44 This is from a Further Maths Examination Paper Mark scheme.
 
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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