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 67 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 2825 14 2842:23 Alcohol (grams) 0.24 ± 0.07 0.39 ± 0.15 x (a) Give and s for each of the four sets of measurements. (Give answers accurate to 3 decimal places.) Drivers Total Calories: 2825 s = 135.735 Drivers Alcohol:24 Conductors Total Calories:2842 5188.263 conductors Alcohol: 139 s 1.228 (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= .831 Conclusion O Reject Ho Do not reject Ho. (c) Hov 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.) Do not reject Ho. (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

you need answer of c. d and e

(c) here we use t-test with null hypothesis H0:mean1=mean2 and alternate hypothesis H1:mean1mean2

t=(mean1-mean2)/((sp*(1/n1 +1/n2)1/2) and sp2=((n1-1)s12+(n2-1)s22)/n and with df is n=n1+n2-2

t=0.99 and df=159

since two-tailed p-value=0.3229 is more than alpha=0.05 so we do not reject H0.

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

95% confidence interval for =0.39±t(0.05/2, 67-1)*0.15=0.39±1.9965*0.15= 0.390±0.0.299=(0.091,0.689)   

(d) SE(difference)=sp*(1/n1 +1/n2)1/2)=0.9461*sqrt(1/94+1/67)=0.1513

and sp2=((n1-1)s12+(n2-1)s22)/n and with df is n=n1+n2-2

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

99% confidence interval for =0.15±t(0.05/2, 159)*0.15=0.39±1.975*0.1232= 0.15±0.243=(-0.087,0.393)

t-test total calories sample mean s s2 n (n-1)s2 drivers 0.24 0.67867518 0.4606 94 42.8358 conductor 0.39 1.227802916 1.5075 67 99.495 difference= 0.15 1.9681 161 142.3308 sp2= 0.895162264 sp= 0.946130152 t= 0.991582301 one tailed p-value= 0.161453965 two tailed p-value= 0.32290793

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