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
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
1 answer
2 answers