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Problem 2. The data below gives the mean price (in cents) of a litre of regular gasoline at self-service filling stations at

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solution:-

part (a) sample mean = 130.37

part (b) sample standard deviation 5.21

part (c)

here degree of freedom df = n - 1 = 6 - 1 = 5

we look into t table with df and with 90% confidence

critical value t = 2.015

confidence interval formula

=> sample mean +/- t * standard deviation/sqrt(n)

=> 130.37 +/- 2.015 * 5.21/sqrt(6)

=> (126.08 , 134.66)


part (d)

here degree of freedom df = n - 1 = 6 - 1 = 5

we look into t table with df and with 95% confidence

critical value t = 2.571

confidence interval formula

=> sample mean +/- t * standard deviation/sqrt(n)

=> 130.37 +/- 2.571 * 5.21/sqrt(6)

=> (124.90 , 135.84)

the confidence interval be wider than the interval found in part(c)

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