Question

The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. The test statistic is

Answer 1

Null hypothesis:
`\sigma^2 \leq 0.003`

Alternative hypothesis:
`\sigma^2 > 0.003`

And for this case the statistic is given by:

`T= (n-1)(s^2)/(\sigma^2_0)`

And replacing we got:

`T =(26-1) (0.06^2)/(0.003)= 30`

Explanation:

We have the following info given:

`n = 26` represent the sample size

`s =0.06` represent the sample deviation

We want to test the following hypothesis:

Null hypothesis:
`\sigma^2 \leq 0.003`

Alternative hypothesis:
`\sigma^2 > 0.003`

And for this case the statistic is given by:

`T= (n-1)(s^2)/(\sigma^2_0)`

And replacing we got:

`T =(26-1) (0.06^2)/(0.003)= 30`