Question

Do various occupational groups differ in their diets? A British study of this question compared 87 drivers and 66 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 2821 ± 14 2850 ± 15 Alcohol (grams) 0.27 ± 0.1 0.35 ± 0.14

(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 =2821

s =14*sqrt(87)=130.58

Drivers Alcohol:

x =0.27

s =0.1*sqrt(87)=0.9327

Conductors Total Calories: x =2850

s =15*sqrt(66)=121.86

Conductors Alcohol: x =0.35

s =0.14*sqrt(66)=1.1374

standard error(se)=s/sqrt(n)

(b)t=1.4,

df=151

Donot reject H0 ( as one tailed p-value is more than alpha=0.05)

here we use t-test with

null hypothesis H0:mean1=mean2 and

alternate hypothesis H1:mean1

statistic t=(mean1-mean2)/((s_p*(1/n1 +1/n2)^1/2) =1.4

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

	sample	mean	s	s2	n	(n-1)s2
	Driver	2821	130.58	17051.1364	87	1466397.73
	conductor	2850	121.86	14849.8596	66	965240.874
	difference=	29		31900.996	153	2431638.604
	sp2=	16103.57
	sp=	126.90
	t=	1.40
one tailed	p-value=	0.0818

(c) t=0.48

df=151,

Donot Reject H0 (p-value is more than typical alpha=0.05)

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) =0.48

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

	sample	mean	s	s2	n	(n-1)s2
	Driver	0.27	0.932737905	0.87	87	74.82
	conductor	0.35	1.137365377	1.2936	66	84.084
	difference=	0.08		2.1636	153	158.904
	sp2=	1.05
	sp=	1.03
	t=	0.48
one tailed	p-value=	0.3168
two tailed	p-value=	0.6335

(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)

n	sample mean	sd
66	0.35	1.14

	t-value	margin of error	lower limit	upper limit
95% confidence interval	1.997	0.280	0.070	0.630

(e) (1-alpha)*100% confidence interval for difference population mean=

=difference sample mean±t(alpha/2,n)*SE(difference of sample mean)

99% confidence interval =0.08±t(0.01/2, 151)*SE(difference)=0.08±1.997*0.1674=0.08±0.3343=(-0.2543,0.4143)

SE(difference)=(s_p*(1/n1 +1/n2)^1/2) =0.1674

please look part (c) also to find the SE(difference)