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 13 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.
1.) Find the value of the chi-square statistic for the sample.
2.) Find the requested confidence interval for the population standard deviation.
lower limit _____min
upper limit _______ min
To determine if Jim is justified in claiming that the variance is larger than stated in his journal, we can conduct a chi-square test for variance.
1.) Find the value of the chi-square statistic for the sample:
Step 1: Define the null and alternative hypotheses:
Null hypothesis (H0): The variance is equal to that stated in Jim's journal.
Alternative hypothesis (Ha): The variance is larger than that stated in Jim's journal.
Step 2: Calculate the chi-square statistic:
The formula to calculate the chi-square statistic for variance is:
χ² = (n - 1) * s² / σ²
Where:
n = sample size
s² = sample variance
σ² = hypothesized population variance
Given:
n = 13
s² = 25² = 625 (since standard deviation = 25)
σ² = 15² = 225 (assuming the stated population standard deviation is correct)
Plugging these values into the formula, we get:
χ² = (13 - 1) * 625 / 225
= 12 * 625 / 225
= 33.33
2.) Find the requested confidence interval for the population standard deviation:
To find the confidence interval for the population standard deviation (σ), we will use the chi-square distribution and the formula:
[ (n - 1) * s² / χ², (n - 1) * s² / χ² ]
Where:
n = sample size
s² = sample variance
χ² = chi-square value corresponding to the desired level of significance and degrees of freedom
Given:
n = 13
s² = 25² = 625 (since standard deviation = 25)
α = 0.05 (for a 95% confidence interval)
degrees of freedom = n - 1 = 13 - 1 = 12
Step 1: Look up the critical chi-square value for α/2 and degrees of freedom in a chi-square distribution table. For α/2 = 0.05/2 = 0.025 and degrees of freedom = 12, the critical chi-square value is approximately 23.34.
Step 2: Calculate the lower and upper limits of the confidence interval:
Lower limit = (n - 1) * s² / χ²
= 12 * 625 / 23.34
≈ 321.31 minutes
Upper limit = (n - 1) * s² / χ²
= 12 * 625 / 23.34
≈ 848.55 minutes
Therefore, the 95% confidence interval for the population standard deviation is:
Lower limit ≈ 321.31 minutes
Upper limit ≈ 848.55 minutes.
X^2 = ∑ (O-E)^2/E, where O = observed frequency and E = expected frequency.
∑ = sum of all the cells.
E = (column total * row total)/grand total
df = n - 1, where n = number of cells
Look up value in X^2 table in the back of your textbook.
95% = mean ± 1.96 SEm
SEm = SD/√n