Question

Do various occupational groups differ in their diets? A British study of this question compared 95 drivers and 69 conductors of London double-decker buses. The conductors' jobs require more physical activity. The article reporting the study gives the data as "Mean daily consumption ± (se)." Some of the study results appear below. Drivers Conductors Total calories 2827 ± 13 2846 ± 20 Alcohol (grams) 0.28 ± 0.07 0.38 ± 0.13 (a) Give x and s for each of the four sets of measurements. (Give answers accurate to 3 decimal places.) Drivers Total Calories: x = s = Drivers Alcohol: x = s = Conductors Total Calories: x = s = Conductors Alcohol: x = s = (b) Is there significant evidence at the 5% level that conductors consume more calories per day than do drivers? Use the conservative two-sample t method to find the t-statistic, and the degrees of freedom. (Round your answer for t to three decimal places.) t = df = Conclusion Reject H0. Do not reject H0. (c) How significant is the observed difference in mean alcohol consumption? Use the conservative two-sample t method to obtain the t-statistic. (Round your answer to three decimal places.) t = Conclusion Reject H0. Do not reject H0. (d) Give a 95% confidence interval for the mean daily alcohol consumption of London double-decker bus conductors. (Round your answers to three decimal places.) ( , ) (e) Give a 99% confidence interval for the difference in mean daily alcohol consumption for drivers and conductors. (conductors minus drivers. Round your answers to three decimal places.) ( , )

Answer 1

a)

Drivers Calories: - 13=s/ $\sqrt$ 95, s=126.71, x=2827

Driver Alcohol: - 0.07=s/ $\sqrt$ 95, s=0.6823, x=0.28

Conductor Calories: - 20=s/ $\sqrt$ 69, s=166.13, x=2846

Conductor Alcohol: - 0.13=s/ $\sqrt$ 69, s=1.08, x=0.38

b)

H0: μ1 – μ2 <= 0

H1: μ1 – μ2 > 0

Sp^2=(n1-1)*S1^2+(n2-1)*S2^2/(n1-1)+(n2-1)

=(69-1)*166.13^2+(95-1)*126.71^2/(69-1)+(95-1)

=20900.95

t_STAT=(X1-X2)-( $\mu$ 1- $\mu$ 2)/ $\sqrt$ Sp^2(1/n1+1/n2)

=(2846-2827)-0/ $\sqrt$ 20900.95(1/69+1/95)

=0.83

t_CRIT is 1.9747 (df=162). Hence, cannot reject the null hypothesis.

c)

H0: μ1 – μ2 = 0

H1: μ1 – μ2 # 0

Sp^2=(n1-1)*S1^2+(n2-1)*S2^2/(n1-1)+(n2-1)

=(69-1)*1.08^2+(95-1)*0.6823^2/(69-1)+(95-1)

=0.7597

t_STAT=(X1-X2)-( $\mu$ 1- $\mu$ 2)/ $\sqrt$ Sp^2(1/n1+1/n2)

=(0.38-0.28)-0/ $\sqrt$ 0.7597(1/69+1/95)

=0.7253

t_CRIT is +/-1.976 (df=162). Hence, cannot reject the null hypothesis.

d)

0.38+/-1.9954*0.13

0.38-0.259402=0.120598 to 0.38+0.259402=0.639402