Question

A new kind of typhoid shot is being developed by a medical research team. The old typhoid shot was known to protect the population for a mean time of 36 months, with a standard deviation of 3 months. To test the time variability of the new shot, a random sample of 29 people were given the new shot. Regular blood tests showed that the sample standard deviation of protection times was 1.9 months. Using a 0.05 level of significance, test the claim that the new typhoid shot has a smaller variance of protection times.

Answer 1

its a.....

Explanation:

Answer 2

`\chi^2 =(29-1)/(9) 3.61 =11.231`

`p_v =P(\chi^2_(28) <11.231)=0.002`

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(11.231,28,TRUE)"

If we compare the p value and the significance level provided we see that
`p_v <\alpha` so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And that means that the population variance is significantly lower than 9 (old variance)

Explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

`n=29` represent the sample size

`\alpha=0.05` represent the confidence level

`s^2 =1.9^2= 3.61` represent the sample variance obtained

represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance specification is violated, so the system of hypothesis would be:

Null Hypothesis:
`\sigma^2 \geq 9`

Alternative hypothesis:
`\sigma^2 <9`

Calculate the statistic

For this test we can use the following statistic:

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

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

`\chi^2 =(29-1)/(9) 3.61 =11.231`

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case df=29-1=28. And since is a left tailed test the p value would be given by:

`p_v =P(\chi^2_(28) <11.231)=0.002`

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(11.231,28,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that
`p_v <\alpha` so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And that means that the population variance is significantly lower than 9 (old variance)