Question

Researchers are interested in the effect of a certain nutrient on the growth rate of plant seedlings. Using a hydroponics grow procedure that utilized water containing the​ nutrient, they planted six tomato plants and recorded the heights of each plant 14 days after germination. Those​ heights, measured in​ millimeters, were 55.7, 60.6, 61.3​, 62.7​, 65.7, and 71.9. Complete parts 1 through 4 below.

1. Using​ technology, find the​ 95% confidence interval for the population mean mu.

(____,_____) Round to one decimal place as needed.

2. Name two things you could do to get a narrower interval than the one in part a. Choose the correct choice below.

a. increase the confidence level or decrease the sample size.

b. decrease the confidence level or decrease the sample size.

c. increase the confidence level or increase the sample size

d. decrease the confidence level or decrease the sample size

3. Using​ technology, construct a​ 99% confidence interval.

(____,____) Round to one decimal place as needed.

Why is the​ 99% confidence interval wider than the​ 95% interval?

a. The t-distribution critical value is smaller with a higher confidence level.

b. The standard deviation is smaller for a higher confidence level.

c. The t-distribution critical value is larger with a higher confidence level.

d. None of the above statements are correct.

4. On what assumptions is the interval in part a​ based?

a. The data are obtained by randomization and the population distribution is approx. uniform.

b. The population distribution is approx. normal and all the data are within two standard deviations of the mean.

c. The data are obtained by randomization and are within two standard deviations of the mean.

d. The data are obtained by randomization and the population distribution is approx. normal.

Answer 1

1)

sample mean, xbar = 62.983
sample standard deviation, s = 5.4503
sample size, n = 6
degrees of freedom, df = n - 1 = 5

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.571

ME = tc * s/sqrt(n)
ME = 2.571 * 5.4503/sqrt(6)
ME = 5.7

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (62.983 - 2.571 * 5.4503/sqrt(6) , 62.983 + 2.571 * 5.4503/sqrt(6))
CI = (57.3 , 68.7)

2)

b. decrease the confidence level or decrease the sample size.

3)

sample mean, xbar = 62.983
sample standard deviation, s = 5.4503
sample size, n = 6
degrees of freedom, df = n - 1 = 5

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 4.032

ME = tc * s/sqrt(n)
ME = 4.032 * 5.4503/sqrt(6)
ME = 9

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (62.983 - 4.032 * 5.4503/sqrt(6) , 62.983 + 4.032 * 5.4503/sqrt(6))
CI = (54.0 , 72.0)

c. The t-distribution critical value is larger with a higher confidence level.

4)

d. The data are obtained by randomization and the population distribution is approx. normal.