Question

Jim Mead is a veterinarian who visits a Vermont farm to examine prize bulls. In order to examine a bull, Jim first gives the animal a tranquilizer shot. The effect of the shot is supposed to last an average of 65 minutes, and it usually does. However, Jim sometimes gets chased out of the pasture by a bull that recovers too soon, and other times he becomes worried about prize bulls that take too long to recover. By reading journals, Jim has found that the tranquilizer should have a mean duration time of 65 minutes, with a standard deviation of 15 minutes. A random sample of 10 of Jim's bulls had a mean tranquilized duration time of close to 65 minutes but a standard deviation of 26 minutes. At the 1% level of significance, is Jim justified in the claim that the variance is larger than that stated in his journal? Find a 95% confidence interval for the population standard deviation.

(a) What is the level of significance?


State the null and alternate hypotheses.

H_o: σ² = 225; H₁: σ² ≠ 225H_o: σ² = 225; H₁: σ² > 225    H_o: σ² = 225; H₁: σ² < 225H_o: σ² > 225; H₁: σ² = 225

(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 binomial population distribution.    We assume a uniform population distribution.We assume a exponential 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 1% level of significance, there is insufficient evidence to conclude that the variance of the duration times of the tranquilizer is larger than stated in the journal.At the 1% level of significance, there is sufficient evidence to conclude that the variance of the duration times of the tranquilizer is larger than stated in the journal.    

(f) Find the requested confidence interval for the population standard deviation. (Round your answers to two decimal place.)

lower limit	min
upper limit	min

Interpret the results in the context of the application.

We are 95% confident that σ lies above this interval.We are 95% confident that σ lies outside this interval.    We are 95% confident that σ lies within this interval.We are 95% confident that σ lies below this interval.

Answer 1

(a)

(i)

The level of significance = = 0.01

(ii)

Correct option:

H_o: σ² = 225; H₁: σ² > 225  

(b)

(i)

The value of the chi-square statistic for the sample is got as follows:

So,

The value of the chi-square statistic for the sample = 27.04

(ii)

the degrees of freedom = n - 1 = 10 - 1 = 9

So,

the degrees of freedom = 9

(iii)

Correct option:

We assume a normal population distribution.

(c)

By Technology, p value = 0.0014

So,

Correct option:

P-value < 0.005

(d)

Correct option:

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

(e)

Correct option:

At the 1% level of significance, there is sufficient evidence to conclude that the variance of the duration times of the tranquilizer is larger than stated in the journal.    

(f)

(i)

n = 10

s² = 26² = 676

= 0.05

lower limit =

upper limit =

So,

Answer is:

lower limit	17.88
upper limit	47.47

(ii) Correct option:

We are 95% confident that σ lies within this interval.