Hypothesis testing

Question

I have been given the data set:

42 36 46 43 41 35 43 45 40 39

The null hypothesis is that $\mu=42$ and alternative hypothesis is that $\mu<42.$

I have found that $s=3.59011, \bar X=41$.

I have then used this to find the test statistic by doing the following:

$$t=\frac{\sqrt{10}(41-42)}{3.59011}=-0.8808$$

Now in order to find the p-value I have found $$\mathbb{P}(T<-0.8808)=\mathbb{P}(T>0.8808)=1-\mathbb{P}(T<0.8808)=1-0.7993325=0.2006675.$$

This all looks correct to me but when I try to find the p-value using R it says that the p-value should be $0.4013$

Any ideas where I'm going wrong?

I am given that we want a significance level, $\alpha=0.05$.

Also, am I right in saying that the critical region $C=\{T<c^*\}$ where $c^*=1.833113$?

From my p-value I wouldn't reject the null hypothesis as my p-value is greater than 0.05. However, it lies within this region so that would say do reject it. Am I missing something?

Answer 1

You have the right p-value, but you did not understand the software. The p-value that you say you got from R is for a two-sided alternative hypothesis that says $\mu\ne42$ instead of $\mu<42$.

Your critical value is correct. [PS: I hadn't noticed until it was pointed out by "heropup" that the minus sign was missing. It would be correct if you put that in.]

Answer 2

Your critical region is incorrect. For a one-sided (left-tailed) test where the alternative hypothesis is that the true mean is less than some value, the critical region would correspond to a $t$-statistic that is less than the lower $100\alpha$ percentile of the student's $t$ distribution with $n-1$ degrees of freedom. Your critical region is way too big because the critical value you have chosen is the wrong sign: it should be $-1.833113$.