Mathematics, 13.03.2020 04:32 restinpeicejahseh

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.

Answers: 3

Show answers

Answers

Answer from: Thejollyhellhound20

$\chi^2 =\frac{29-1}{9} 3.61 =11.231$

$p_v =P(\chi^2_{28}$

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 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)

Step-by-step 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: $\sigma^2$

Calculate the statistic

For this test we can use the following statistic:

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

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

$\chi^2 =\frac{29-1}{9} 3.61 =11.231$

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:

$p_v =P(\chi^2_{28}$

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

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

Conclusion

Answer from: Quest

they are simplified

step-by-step explanation:

Answer from: Quest

She should use 12/3 to measure

Another question on Mathematics

Mathematics, 21.06.2019 19:00

Acompany that manufactures and sells guitars made changes in their product range. from the start they had 20 models. then they reduced the number of models to 15. as a result, the company enjoyed a 10% increase in turnover. - how much did the stock level change? in (%)

Answers: 2

Answer

Mathematics, 21.06.2019 19:00

There is an entrance to the computer room at point e which lies on ab and is 5 feet from point a. plot point e on the coordinate plane. find the distance from the entrance at point e to the printer at point e

Answers: 1

Answer

Mathematics, 21.06.2019 22:30

In order to be invited to tutoring or extension activities a students performance 20 point higher or lower than average for the

Answers: 1

Answer

Mathematics, 21.06.2019 22:30

Solve for x. −3/4(x+2)=6 enter your answer in the box

Answers: 1

Answer

You know the right answer?

A new kind of typhoid shot is being developed by a medical research team. The old typhoid shot was k...