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# 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.

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#### Solution Preview

Sample size= 30
within plus /minus \$0.03

Mean=&#956;= \$1.46
Standard deviation =&#963;= \$0.15
sample size=n= 30
&#963;x=standard error of mean=&#963;/&#8730;n= \$0.0274 = ( 0.15 /&#8730; 30)
x1 bar = \$1.43 =1.46+0.03
x2bar= \$1.49 =1.46-0.03
z1=(x1bar -&#956;)/&#963;x= -1.0949 =(1.43-1.46)/0.0274
z2=(x2bar-&#956;)/&#963;x= 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%

Sample size= 50
within plus /minus \$0.03

Mean=&#956;= \$1.46
Standard deviation =&#963;= \$0.15
sample size=n= 50
&#963;x=standard error of mean=&#963;/&#8730;n= \$0.0212 = ( 0.15 /&#8730; 50)
x1 bar = \$1.43 =1.46+0.03
x2bar= \$1.49 =1.46-0.03
z1=(x1bar -&#956;)/&#963;x= -1.4151 =(1.43-1.46)/0.0212
z2=(x2bar-&#956;)/&#963;x= 1.4151 =(1.49-1.46)/0.0212

Cumulative Probability corresponding ...

#### Solution Summary

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

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