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 6 of Jim's bulls had a mean tranquilized duration time of close to 65 minutes but a standard deviation of 27 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? 0.05 State the null and alternate hypotheses. O Ho: 02 > 225; H,: 02 = 225 OH:02 = 225; H7:< 225 OH:02 = 225; H,: 02 + 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. OP-value > 0.100 O 0.050 < P-value < 0.100 0.025 < P-value < 0.050 0.010 < P-value < 0.025 O 0.005 < P-value < 0.010 OP-value < 0.005

Answer 1

b)

test statistic X2 =(n-1)s2/ σ2=(6-1)*(27/15)^2 =

16.20

degree of freedom =n-1=5

We assume a normal population distribution.

c)

O 0.005 < P-value < 0.010

Since the P-value sa, 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)

	here n =	6
	s2=	729.000
Critical value of chi square distribution for n-1=5 df and 95 % CI
Lower critical value χ2L=		0.831
Upper critical valueχ2U=		12.833

for Confidence interval of standard deviation:
Lower bound =√((n-1)s2/χ2U)=		16.85
Upper bound =√((n-1)s2/χ2L)=		66.22

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