Yes as others have already mentioned, it's important to note that, regardless of whether it's a one-tail or two-tail test, your G.C. will always churn out the *total* p-value against the 'total' significant level.

So when you key in the following for your 2-tail test at 10% significance level as shown below

the calculated p-value of 0.09487 is compared against 10% significance level.

As the distribution is symmetrical, you may also look at this as effectively comparing a p-value of 0.047 against 5% at each end as shown below:

]]>Have a look at my comment here. You used `y` = √[`x`^{2}(`x`+2)] to sketch your f'(`x`)?

i get it now, thank you so much 🙂

]]>For Q3(ii)(b), did you do this on your G.C.?

And sketched this?

Do note that to obtain the graph of `y`^{2} = `x`^{2}(`x` + 2), you'll need to plot

`y` = ±`x`√(`x` + 2) which consists of the graph of `y` = `x`√(`x` + 2) AND the graph of `y` = −`x`√(`x` + 2).

The correct way of using your G.C. would be to plot the curves for `y` = `x`√(`x` + 2) and `y` = − `x`√(`x` + 2) as follows:

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**Min Han** is right. See my little write-up on continuity correction.

Already updated in version 1.3 of the solutions. Sorry to miss out your question earlier - it's not easy trawling through this pile of comments 🙁

]]>Thanks for your feedback! I've added a reference to the non-linear relationship from part (i) for Q10(iii) in version 1.3 of the solutions!

]]>For Q3(iii), we're sketching the gradient function of f(`x`) i.e. f'(`x`) *not* the inverse function of f(`x`) i.e. f^{−1}(`x`)! You are not alone in this one *sigh*

For Q7(v), would these diagrams help?

For Q11(iii) & (iv), have a look at this comment.

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