problem with the chi-square table while solving hypothesis test i was looking up the chi-square table for n=4, and 0.98 probability but i couldn't find a column of 0.98 so what should i do to find out the value? 
The question is to find out if the null hypothesis will be rejected.
H0 : variance = 2 , Ha : variance < 2.  
sample variance = 0.716 
sample size = 5
type 1 error probability = 0.02
and that's why i need to find out 0.98 with df=4 in the table. Please help.
 A: I just realized you have sample size of $5$ and $4$.. Usually sample size is denoted by n so that is confusing for the viewers. if the sample size is $5$, then the degrees of freedom (df) is $4$, which in this case:
 qchisq(.98,4) = 11.67

Confidence Interval
You could create a 98% confidence interval estimate of the variance as follows:
I'm doing to define these first since I don't know how to do subscripts in here, so let
$$X^2 \left(1-\frac{\alpha }{2}\right) = X^2(.99)$$
$$and$$
$$X^2 \left(\frac{\alpha }{2}\right) = X^2(.01)$$
here we used df = 4...
Then, using a chi square table, I obtain the values:
$$X^2(.99) = 13.28$$
$$and$$
$$X^2(.01) = .297$$
The 98% CI for the estimate of the variance is given by:
$$\frac{(n-1) s^2}{X^2(0.99)}\leq \sigma ^2\leq \frac{(n-1) s^2}{X^2(.01)}$$
Thus, 
$$\frac{(5-1) 0.716}{13.28}\leq \sigma ^2\leq \frac{(5-1) 0.716}{.297}$$
The 98% CI for the estimate of the variance is:
$$.2156 \leq \sigma ^2 \leq 9.64$$
$$and$$
$$.464 \leq \sigma \leq 3.105$$
Hypothesis Test
So notice that since the alternative hypothesis is "<", this means we will be conducting a lower (one sided) hypothesis test, meaning our critical value will be X^2(.01) = .297, and if our observed chi statistic exceeds this value, then the Null is not rejected
We can compute the observed Chi Square statistic based on the data you have using the formula:
$$X^2=\frac{(n-1) s^2}{\sigma ^2}$$
Using those $3$ values I computed a chi square statistic of $1.43$
Since $1.43$ > $.297$, the null hypothesis is not rejected.. Therefore, there is not enough evidence to claim that the population variance  is less than 2, at the .02 significance level.
Phew!!
