Question

If the quality of teaching is similar in a school, the scores on a standardized test will have a standard deviation of 12. The superintendent wants to know if there is a disparity in teaching quality, and decides to investigate whether the standard deviation of test scores has changed. She samples 25 random students and finds a mean score of 156 with a standard deviation of 5.4822. Is there evidence that the standard deviation of test scores has decreased at the α=0.1 level? Assume the population is normally distributed.

Answer 1

Answer:

$p_v =P(\chi^2 <5.009)=0.0000128$

We can use the following excel code to find the p value for example

"=CHISQ.DIST(5.009,24,TRUE)"

We can see that the p value is much lower than the significance level given of 0.1 so then we have enough evidence to reject the null hypothesis, so we can conclude that the true deviation is significantly lower than 12 at 10% of significance.

Step-by-step explanation:

Data given

$n=25$ represent the sample size selected

$\alpha=0.1$ represent the confidence level  

$s^2 =5.4822^2 = 30.055$ represent the sample variance obtained

$\sigma^2_0 =12^2= 144$ represent the value that we want to test

System of hypothesis

On this case we want to check if the population deviation is less than 12, we can check this using the test in terms of the variance, the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 144$

Alternative hypothesis: $\sigma^2 <144$

Calculate the statistic  

The statistic for this test is given by:

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

The degrees of freedom are :

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

$\chi^2 =\frac{25-1}{144} 5.4822^2 =5.009$

Statistical decision

We have a left tailed test so the p value would be given by:

$p_v =P(\chi^2 <5.009)=0.0000128$

We can use the following excel code to find the p value for example

"=CHISQ.DIST(5.009,24,TRUE)"

We can see that the p value is much lower than the significance level given of 0.1 so then we have enough evidence to reject the null hypothesis, so we can conclude that the true deviation is significantly lower than 12 at 10% of significance.