1 The American Auto Association reports the mean price per gallon of regular gas

Question

1 The American Auto Association reports the mean price per gallon of regular gasoline is $3.10, with a population standard deviation of $0 20. Assume a random sample of 16 gasoline stations is selected and their mean cost for regular gasoline is computed a. What is the standard error of the mean in this experiment? b. What is the probability that the sample mean is between $2.98 and $3.127 c What is the probability that the difference between the sample mean and the population mean is less than 0.01? d. What is the likelihood the sample mean is greater than $3.08?

Explanation / Answer

population mean = 3.10

population standard deviation = 0.20

sample size i.e n= 16

a)

standard error of sample of size n = /n

standard error of mean of the experiment = 0.20/16 = 0.20/4 = 0.05

b)

z value = (X-mean)/standard deviation

z value for sample mean = 2.98 is (2.98-3.10)/0.05 = -2.4, corresponding p value using z table is 0.0082

P(sample mean <= 2.98) = 0.0082

z value for sample mean = 3.12 is (3.12-3.10)/0.05 = 0.4, corresponding p value using z table is 0.6554

P(sample mean <= 3.12) = 0.6554

P(2.98 <= sample mean <= 3.12) = P(sample mean <= 3.12) - P(sample mean <= 2.98)

= 0.6554 -0.0082 = 0.6472

probability is sample mean is between 2.98 and 3.12 is 0.6472

c)

probability that the difference between the sample mean and the populatio mean is less than 0.01 = probability that the difference between the sample mean between (3.10-0.01, 3.10+0.01) i.e (3.09, 3.11)

z value for sample mean = 3.09 is (3.09-3.10)/0.05 = -0.2, corresponding p value using z table is 0.4207

P(sample mean <= 3.09) = 0.4207

z value for sample mean = 3.11 is (3.11-3.10)/0.05 = 0.2, corresponding p value using z table is 0.5793

P(sample mean <= 3.11) = 0.5793

P(3.09 <= sample mean <= 3.11) = P(sample mean <= 3.11) - P(sample mean <= 3.09)

= 0.5793 -0.4207 = 0.1586

probability that the difference between the sample mean and the populatio mean is less than 0.01 = 0.1586

d)

z value for sample mean = 3.08 is (3.08-3.10)/0.05 = -0.4, corresponding p value using z table is 0.3446

P(sample mean <= 3.08) = 0.3446 => P(sample mean > 3.08) = 1 - 0.3446 = 0.6554

likelohood that the sample mean is greater than 3.08 is 0.6554

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