Question

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

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 23 people were given the new shot. Regular blood tests showed that the sample standard deviation of protection times was 2.1 months. Using a 0.05 level of significance, test the claim that the new typhoid shot has a smaller variance of protection times.

(a) What is the level of significance?


State the null and alternate hypotheses.

Ho: σ2 = 9; H1: σ2 < 9Ho: σ2 < 9; H1: σ2 = 9    Ho: σ2 = 9; H1: σ2 ≠ 9Ho: σ2 = 9; H1: σ2 > 9


(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)


What are the degrees of freedom?


What assumptions are you making about the original distribution?

We assume a normal population distribution.We assume a uniform population distribution.    We assume a exponential population distribution.We assume a binomial population distribution.


(c) Find or estimate the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005


(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.


(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is insufficient evidence to conclude that the new typhoid shot has a smaller variance of protection times.At the 5% level of significance, there is sufficient evidence to conclude that the new typhoid shot has a smaller variance of protection times.    


(f) Find a 90% confidence interval for the population standard deviation. (Round your answers to two decimal places.)

lower limit
upper limit    


Interpret the results in the context of the application.

We are 90% confident that σ lies above this interval.We are 90% confident that σ lies outside this interval.    We are 90% confident that σ lies within this interval.We are 90% confident that σ lies below this interval.

Homework Answers

Answer #1

a) level of significance =0.05

Ho: σ2 = 9; H1: σ2 < 9

b)

value of the chi-square statistic =(23-1)*(2.1/3)2 =10.78

degrees of freedom =23-1=22

We assume a normal population distribution.

c) 0.010 < P-value < 0.025

d)

  Since the P-value ≤ α, we reject the null hypothesis

e)

At the 5% level of significance, there is sufficient evidence to conclude that the new typhoid shot has a smaller variance of protection times.    

f)

lower limit =1.69

upper limit =2.80

  We are 90% confident that σ lies within this interval.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A new kind of typhoid shot is being developed by a medical research team. The old...
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 19 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...
A new kind of typhoid shot is being developed by a medical research team. The old...
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 25 people were given the new shot. Regular blood tests showed that the sample standard deviation of protection times was 1.7 months. Using a 0.05 level of significance, test...
A new kind of typhoid shot is being developed by a medical research team. The old...
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 25 people were given the new shot. Regular blood tests showed that the sample standard deviation of protection times was 1.5 months. Using a 0.05 level of significance, test...
A new kind of typhoid shot is being developed by a medical research team. The old...
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 23 people were given the new shot. Regular blood tests showed that the sample standard deviation of protection times was 1.7 months. Using a 0.05 level of significance, test...
Let x = age in years of a rural Quebec woman at the time of her...
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 31 women in rural Quebec gave a sample variance s2 = 2.5. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence...
Let x = age in years of a rural Quebec woman at the time of her...
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 51 women in rural Quebec gave a sample variance s2 = 3.4. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence...
Let x = age in years of a rural Quebec woman at the time of her...
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 41 women in rural Quebec gave a sample variance s2 = 3.1. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence...
Let x = age in years of a rural Quebec woman at the time of her...
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 51 women in rural Quebec gave a sample variance s2 = 2.8. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence...
Let x = age in years of a rural Quebec woman at the time of her...
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 31 women in rural Quebec gave a sample variance s2 = 2.4. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence...
Let x = age in years of a rural Quebec woman at the time of her...
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 51 women in rural Quebec gave a sample variance s2 = 3.5. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence...