Do various occupational groups differ in their diets? A British study of this qu
ID: 3356154 • Letter: D
Question
Do various occupational groups differ in their diets? A British study of this question compared 93 drivers and 68 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.)
(____________ , _________)
Drivers Conductors Total calories 2828 ± 12 2849 ± 23 Alcohol (grams) 0.29 ± 0.12 0.37 ± 0.15Explanation / Answer
Part a
We have,
S/sqrt(n) = se
S = se*sqrt(n)
Required values of standard deviations are given as below:
n
93
68
Drivers
Conductors
Xbar
Se
S
Xbar
Se
S
Total Calories
2828
12
12*sqrt(93) = 115.7238
2849
23
23*sqrt(68) = 189.6629
Alcohol
0.29
0.12
0.12*sqrt(93) = 1.157238
0.37
0.15
0.15*sqrt(68) = 1.236932
Drivers Total Calories: x = 2828
s = 115.724
Drivers Alcohol: x = 0.29
s = 1.157
Conductors Total Calories: x = 2849
s = 189.663
Conductors Alcohol: x = 0.37
s = 1.2369
Part b
Here, we have to use the two sample t test for the population means. Formulas and calculations are given as below:
H0: µ1 = µ2
Versus
Ha: µ1 > µ2
µ1 = Mean for conductor
µ2 = Mean for driver
Degrees of freedom = n1 + n2 – 2 = 68 + 93 – 2 = 159
Test statistic = t = (X1bar – X2bar) / sqrt[Sp2 ((1/n1)+(1/n2))]
Where, Sp2 = pooled variance
Sp2 = (n1*S1^2 + n2*S2^2) / (n1 + n2)
Calculations are carried out by using excel due to lengthy calculations.
Results are given as below:
Data
Hypothesized Difference
0
Level of Significance
0.05
Population 1 Sample
Sample Size
68
Sample Mean
2849
Sample Standard Deviation
189.663
Population 2 Sample
Sample Size
93
Sample Mean
2828
Sample Standard Deviation
115.724
Intermediate Calculations
Population 1 Sample Degrees of Freedom
67
Population 2 Sample Degrees of Freedom
92
Total Degrees of Freedom
159
Pooled Variance
22906.8909
Standard Error
24.1491
Difference in Sample Means
21.0000
t Test Statistic
0.870
Upper-Tail Test
Upper Critical Value
1.6545
p-Value
0.1929
Do not reject the null hypothesis H0
Part c
Here, we have to use same two sample t test as used above.
H0: µ1 = µ2
Versus
Ha: µ1 > µ2
µ1 = Mean for conductor
µ2 = Mean for driver
Degrees of freedom = n1 + n2 – 2 = 68 + 93 – 2 = 159
Test statistic = t = (X1bar – X2bar) / sqrt[Sp2 ((1/n1)+(1/n2))]
Where, Sp2 = pooled variance
Sp2 = (n1*S1^2 + n2*S2^2) / (n1 + n2)
Results by using excel are given as below:
Data
Hypothesized Difference
0
Level of Significance
0.05
Population 1 Sample
Sample Size
68
Sample Mean
0.37
Sample Standard Deviation
1.237
Population 2 Sample
Sample Size
93
Sample Mean
0.29
Sample Standard Deviation
1.157
Intermediate Calculations
Population 1 Sample Degrees of Freedom
67
Population 2 Sample Degrees of Freedom
92
Total Degrees of Freedom
159
Pooled Variance
1.4194
Standard Error
0.1901
Difference in Sample Means
0.0800
t Test Statistic
0.421
Upper-Tail Test
Upper Critical Value
1.6545
p-Value
0.3372
Do not reject the null hypothesis
Part d
We have to find 95% confidence interval for mean daily alcohol consumption for conductors.
Confidence interval = Xbar -/+ t*S/sqrt(n)
We are given
Sample size = n = 68
Xbar = 0.37
S = 1.237
df = n – 1 = 67
Critical value = t = 1.9960
Confidence interval = 0.37 -/+ 1.9960* 1.237/sqrt(68)
Confidence interval = 0.37 -/+ 0.2994
Lower limit = 0.37 - 0.2994 = 0.071
Upper limit = 0.37 + 0.2994 = 0.669
Confidence interval = (0.071, 0.669)
Part e
Here, we have to find 99% confidence interval for difference between the alcohol consumption for drivers and conductors.
Confidence interval = (X1bar – X2bar) -/+ t*sqrt[Sp2 ((1/n1)+(1/n2))]
Where, Sp2 = pooled variance
Sp2 = (n1*S1^2 + n2*S2^2) / (n1 + n2)
Calculations by using excel are given as below:
Data
Hypothesized Difference
0
Level of Significance
0.05
Population 1 Sample
Sample Size
68
Sample Mean
0.37
Sample Standard Deviation
1.237
Population 2 Sample
Sample Size
93
Sample Mean
0.29
Sample Standard Deviation
1.157
Intermediate Calculations
Population 1 Sample Degrees of Freedom
67
Population 2 Sample Degrees of Freedom
92
Total Degrees of Freedom
159
Pooled Variance
1.4194
Standard Error
0.1901
Difference in Sample Means
0.0800
Confidence level = 99%
Critical t value = 2.6071
Margin of error = t*sqrt[Sp2 ((1/n1)+(1/n2))] = 2.6071*sqrt[1.4194*(1/68)+(1/93)]
Margin of error = 0.4956
(X1bar – X2bar) = 0.37 – 0.29 = 0.08
Confidence interval = 0.08 -/+ 0.4956
Lower limit = 0.08 – 0.4956 = -0.416
Upper limit = 0.08 + 0.4956 = 0.576
n
93
68
Drivers
Conductors
Xbar
Se
S
Xbar
Se
S
Total Calories
2828
12
12*sqrt(93) = 115.7238
2849
23
23*sqrt(68) = 189.6629
Alcohol
0.29
0.12
0.12*sqrt(93) = 1.157238
0.37
0.15
0.15*sqrt(68) = 1.236932
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