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 7 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.

Ho: σ2 = 225; H1: σ2 < 225

Ho: σ2 > 225; H1: σ2 = 225

Ho: σ2 = 225; H1: σ2 ≠ 225

Ho: 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 exponential population distribution.

We assume a uniform population distribution.

We assume a binomial population distribution.

We assume a normal population distribution.

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

P-value > 0.100

0.050 < P-value < 0.100

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

P-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 below this interval.

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

Answer 1

Ans:

a) level of significance=0.01

H_o: σ² = 225; H₁: σ² > 225

b)

chi-square statistic=(7-1)*26^2/15^2=18.03

degrees of freedom=7-1=6

We assume a normal population distribution.

c)

p-value=CHIDIST(18.03,6)=0.0062

0.005 < P-value < 0.010

d)

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

e)

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)alpha=1-0.95=0.05

alpha/2=0.025

critical chi square values:

CHIINV(0.025,6)=14.449

CHIINV(0.975,6)=1.237

95% confidence interval for the population standard deviation:

lower limit=SQRT((7-1)*26^2/14.449)=16.75

upper limit=SQRT((7-1)*26^2/1.237)=57.25

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