Question

Do various occupational groups differ in their diets? A British study of this question compared 90 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	2829 ± 13	2847 ± 18
Alcohol (grams)	0.26 ± 0.12	0.37 ± 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) standard error se=s/sqrt(n)

there are 90 drivers and 63 conductor

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

x = 2829

s = 13*sqrt(90)=123.329
Drivers Alcohol:

x =0.26

  s = 0.12*sqrt(90)=1.138
Conductors Total Calories:

x = 2847

s = 18*sqrt(63)=142.871
Conductors Alcohol:

x = 0.37

s = 0.13*sqrt(63)=1.032

(b)t=0.8

df=151

Donot reject H0

here we use t-test with null hypothesis H0:mean1=mean2 and alternate hypothesis H1:mean1< mean2

statistic t=|(mean1-mean2)|/((s_p*(1/n1 +1/n2)^1/2) =0.8

and s_p²=((n1-1)s1²+(n2-1)s2²)/n and with df is n=n1+n2-2=90+63-2=151

	sample	mean	s	s2	n	(n-1)s2
	Driver	2829	123.3288287	15210	90	1353690
	conductor	2847	142.871	20412.12264	63	1265551.604
	difference=	18		35622.12264	153	2619241.604
	sp2=	17345.97
	sp=	131.70
	t=	0.83
one tailed	p-value=	0.2034
two tailed	p-value=	0.4067

since one tailed p-value is more than alpha=0.05

(c)

t=0.61

df=151

Donot reject H0

here we use t-test with null hypothesis H0:mean1=mean2 and alternate hypothesis H1:mean1?mean2

statistic t=(mean1-mean2)/((s_p*(1/n1 +1/n2)^1/2) and s_p²=((n1-1)s1²+(n2-1)s2²)/n and with df is n=n1+n2-2

	sample	mean	s	s2	n	(n-1)s2
	Driver	0.26	1.138	1.295044	90	115.258916
	conductor	0.37	1.032	1.065024	63	66.031488
	difference=	0.11		2.360068	153	181.290404
	sp2=	1.20
	sp=	1.10
	t=	0.61
one tailed	p-value=	0.2710
two tailed	p-value=	0.5420

since two tailed p-value is more than alpha=0.05 ( level of significance =1-confidence), so we accept H0 or donot reject H0

(d) (1-alpha)*100% confidence interval for population mean=sample mean±t(alpha/2,n-1)*s_d/sqrt(n)

95% confidence interval for population mean=mean±t(0.05/2, n-1)*s_d/sqrt(n)=0.37±1.999*1.03/sq(63)=

=0.37±0.260=(0.110,0.630)

n	sample mean	sd
63	0.37	1.03

	t-value	margin of error	lower limit	upper limit
95% confidence interval	1.999	0.260	0.110	0.630

(e)(1-alpha)*100% confidence interval for population difference=sample difference±t(alpha/2,n)*SE(difference)

99% confidence interval =0.11±t(0.01/2,151 )*0.18=0.11±1.976*0.18=0.110±0.356=(-0.246,0.466)

SE(difference)=(s_p*(1/n1 +1/n2)^1/2) =0.18and s_p²=((n1-1)s1²+(n2-1)s2²)/n and with df is n=n1+n2-2

please look in part (c) for SE(difference)