A random sample of the price of gasoline from 30 gas stations in a region gives

Question

A random sample of the price of gasoline from 30 gas stations in a region gives the statistics below. Complete parts a) through c). y-34.39, s-30.25 a) Find a 95% confidence interval for the mean price of regular gasoline in that region. (4297.4.483) Round to three decimal places as needed.) b) Find the 90% confidence interval for the mean. 4.312.4.468) (Round to three decimal places as needed.) c) If we had the same statistics from a sample of 60 stations, what would the 95% confidence interval be now? Round to three decimal places as needed.)

Explanation / Answer

mean is 4.39 and s is 0.25

a) for sample size of 30, standard error SE is s/sqrt(N)=0.25/sqrt(30)=0.04564

z for 95% confidence is 1.96

thus lower bound is mean-z*SE=4.39-1.96*0.04564=4.3005

upper bound is mean+z*SE=4.39+1.96*0.04564=4.4795

b) z for 90% confidence is 1.65

lower bound is mean-z*SE=4.39-1.65*0.04564=4.3147

upper bound is mean+z*SE=4.39+1.65*0.04564=4.4653

c) s for sample size of 60 is se=s/sqrt(N)=0.25/sqrt(60)=0.032275

thus for 95% z is 1.96 and lower bound is mean-z*SE=4.39-1.96*0.32275=3.7574

upper bound is mean+z*SE=4.39+1.96*0.32275=5.02259

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