# Sample size, Sampling distribution of proportions

25) Money magazine reported that the average price of gallon of gasoline in the United States during the first quarter of 2001 was $ 1.46. Assume the price reported by Money is the population mean, and the population standard deviation is $ 0.15.

a) What is the probability that the mean price for a sample of 30 gas stations is within $ 0.03 of the population mean?

b) What is the probability that the mean price for a sample of 50 gas stations is within $ 0.03 of the population mean?

c) What is the probability that the mean price for a sample of 100 gas stations is within $ 0.03 of the population mean?

d) Would you recommend a sample size of 30, 50 or 100 to have at least a 0.95 probability that the sample mean is within $ 0.03 of the population mean?

37.The Food Marketing Institute shows that 17% of households spend more than $100 per week on groceries (U.S.A Today, June 21, 1994). Assume the population proportion is p=0.17 and a simple random sample of 800 households will be selected from the population.

a. Show the sampling distribution of p, the sample proportion of households spending more than $100 per week on groceries.

b. What is the probability that the sample proportion will be within plus/minus 0.02 of the population proportion?

c. Answer part (b) for a sample of 1600 households.

Please see attachment file for details

#### Solution Preview

Sample size= 30

within plus /minus $0.03

Mean=μ= $1.46

Standard deviation =σ= $0.15

sample size=n= 30

σx=standard error of mean=σ/√n= $0.0274 = ( 0.15 /√ 30)

x1 bar = $1.43 =1.46+0.03

x2bar= $1.49 =1.46-0.03

z1=(x1bar -μ)/σx= -1.0949 =(1.43-1.46)/0.0274

z2=(x2bar-μ)/σx= 1.0949 =(1.49-1.46)/0.0274

Cumulative Probability corresponding to z1= -1.0949 is= 0.1368 0r= 13.68%

Cumulative Probability corresponding to z2= 1.0949 is= 0.8632 0r= 86.32%

Therefore probability that the value of x will be between x1bar= $1.43 and x2 bar= $1.49

is = 72.64% =86.32%-13.68%

Answer: 72.64% or 0.7264

Sample size= 50

within plus /minus $0.03

Mean=μ= $1.46

Standard deviation =σ= $0.15

sample size=n= 50

σx=standard error of mean=σ/√n= $0.0212 = ( 0.15 /√ 50)

x1 bar = $1.43 =1.46+0.03

x2bar= $1.49 =1.46-0.03

z1=(x1bar -μ)/σx= -1.4151 =(1.43-1.46)/0.0212

z2=(x2bar-μ)/σx= 1.4151 =(1.49-1.46)/0.0212

Cumulative Probability corresponding ...

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Answers to 2 questions on Sample size, Sampling distribution of proportions