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 9 of Jim's bulls had a mean tranquilized duration time of close to 65 minutes but a standard deviation of 25 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 normal population distribution. We assume a exponential population distribution. We assume a uniform population distribution. We assume a binomial 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.

Answer 1

(a) What is the level of significance?

State the null and alternate hypotheses

Ho: σ2 = 225; H1: σ2 > 225

(b) Find the value of the chi-square statistic for the sample

Test Statistic :-

What are the degrees of freedom

DF = n - 1 = 9 - 1 = 8

What assumptions are you making about the original distribution?

We assume a normal population distribution.

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

P value = P ( > 22.2222 )
P value = 0.0045

P-value < 0.005

(d) Based on your answers

Reject null hypothesis if P value <
Since P value = 0.0045 < 0.01, hence we reject the null hypothesis
Conclusion :- We Reject H0

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

Conclusion :- Accept Alternative Hypothesis

σ2 > 225

There is sufficient evidence to support the claim that the variance is larger than that stated in his journal.