Question

Do various occupational groups differ in their diets? A British study of this question compared 91 drivers and 63 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 ± 14	2843 ± 18
Alcohol (grams)	0.24 ± 0.11	0.37 ± 0.15

(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) Give x and s for each of the four sets of measurements. (Give answers accurate to 3 decimal places.)

Drivers Total Calories: x = 2827
s = 133.551
Drivers Alcohol: x = 0.24
s = 2.289
Conductors Total Calories: x =2843
s = 142.871
Conductors Alcohol: x = 0.37
s = 1.191

Explanation for Value of s

Standard error

Therefore,

For example for drivers total calories:

(b) t = 0.702 Explanation:
df = 63-1 = 62 Conclusion: Do not reject H0 Explanation: Conservative df = minimum of n-1 = minimum of 63-1 and 91-1

p value for t = 0.702 and df =62 is 0.243, since p > 0.05, null hypothesis cannot be rejected.

(c)

t = 0.699,

df = 62

Conclusion: Do not reject H0

(d)

95% confidence interval is (0.084, 0.664)

(e)

99% Confidence interval is [-0.690, 0.950].

Explanation

(X₁ - X₂) ± t*s_{(x1 - x2)} = 0.13 ± (2.61 * 0.31) = 0.13 ± 0.8203

Where,