Question

In a school district, all sixth grade students take the same standardized test. The superintendant of the school district takes a random sample of 25 scores from all of the students who took the test. She sees that the mean score is 134 with a standard deviation of 6.0547. The superintendant wants to know if the standard deviation has changed this year. Previously, the population standard deviation was 13. Is there evidence that the standard deviation of test scores has decreased at the α=0.01 level? Assume the population is normally distributed.

Answer 1

Answer:

$\chi^2 =\frac{25-1}{169} 6.0547^2 =5.206$

$p_v =P(\chi^2$

Since the p value is very low compared with the significance level provided 0.01, we have enough evidence to conclude that the true variance and on this case the deviation is significantly lower than 13 at 1% of significance.

Step-by-step explanation:

We have the following data given

$n=25$ represent the sample size selected for the test

$\alpha=0.01$ represent the confidence level  

$s^2 =6.0547^2 =36.659$ represent the sample variance obtained

$\sigma^2_0 =13^2 = 169$ represent the value that we want to test

System of hypothesis

We want to review if the population deviation is significantly lower than 13 and we can check this with the variance, the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 169$

Alternative hypothesis: $\sigma^2$

Statistic

To test this hypothesis the statistic is:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

The degrees of freedom for this case are:

$df= n-1= 25-1= 24$

Replacing in the statistic formula we got:

$\chi^2 =\frac{25-1}{169} 6.0547^2 =5.206$

P value

since we have a left tailed test the p value would be givne by:

$p_v =P(\chi^2$

We can find the p value with this excel code:

"=CHISQ.DIST(5.206,24,TRUE)"

Since the p value is very low compared with the significance level provided 0.01, we have enough evidence to conclude that the true variance and on this case the deviation is significantly lower than 13 at 1% of significance.