Do various occupational groups differ in their diets? A British study of this qu

Question

Do various occupational groups differ in their diets? A British study of this question compared 94 drivers and 62 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.

(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.)

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.)
(  ,  )

Explanation / Answer

Solution:

Part a

From the given table

n for driver = 94

n for conductor = 62

Drivers Total Calories: Xbar = 2822

SE = S/sqrt(n) = 14

S = 14*sqrt(n) = 14*sqrt(94) = 14* 9.69536 = 135.735

Drivers Alcohol: Xbar = 0.24

SE = 0.06

S/sqrt(n) = 0.06

S = 0.06*sqrt(n) = 0.06*sqrt(94) = 0.06* 9.69536 = 0.582

Conductors Total Calories: Xbar = 2840

S = SE*sqrt(n) = 17*sqrt(62) = 133.8581

Conductors Alcohol: Xbar = 0.38

S = SE*sqrt(n) = 0.12*sqrt(62) = 0.944881

Part b

Here, we have to use two sample t tests for population means.

H0: µ1 = µ2 Vs Ha: µ1 < µ2

(Upper tailed or right tailed test, one tailed test)

= 0.05

Test statistic formula is given as below:

t = (X1bar – X2bar) / sqrt[Sp2 (1/N1)+(1/N2)] where

Sp2 = [(N1 – 1)S1^2 + (N2 – 1)S2^2]/[N1 + N2 – 2]

From the given data for samples, we have

X1bar = 2822

X2bar = 2840

S1 = 135.735

S2 = 133.858

N1 = 94

N2 = 62

DF = N1 + N2 – 2 = 94 + 62 – 2 = 154

Sp2 = [(94 – 1)* 135.735^2 + (62 – 1)* 133.858^2]/[94 + 62 – 2]

Sp2 = 18223.5513

t = (2822 - 2840) / sqrt(18223.5513*((1/94)+(1/62)))

t = -0.8150

Critical values = 1.6548 (By using t-table)

P-value = 0.2082 (by using t-table)

P-value > = 0.05

So, we do not reject the null hypothesis.

Do not reject H0.

There is insufficient evidence that conductors consume more calories per day than do drivers.

Part c

H0: µ1 = µ2 Vs Ha: µ1 < µ2

DF = N1 + N2 – 2 = 94 + 62 – 2 = 154

Sp2 = [(94 – 1)* 0.582^2 + (62 – 1)* 0.945^2]/[94 + 62 – 2]

Sp2 = 0.5583

t = (0.24 – 0.38) / sqrt(0.5583*((1/94)+(1/62)))

t = -1.1452

P-value = 0.1269 (by using t-table)

P-value > = 0.05

Do not reject H0.

There is insufficient evidence that conductors consume more alcohol per day than do drivers.

Part d

Confidence interval = Xbar -/+ t*S/sqrt(n)

Confidence interval = 0.38 -/+ t*0.945/sqrt(62)

We have confidence level = 95% and df = 62 – 1 = 61

So, t = 1.9996 (by using t-table)

Confidence interval = 0.38 -/+ 1.9996*0.945/sqrt(62)

Confidence interval = 0.38 -/+ 0.2400

Lower limit = 0.38 – 0.2400 = 0.14

Upper limit = 0.38 + 0.2400 = 0.62

Confidence interval = (0.140, 0.620)

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.