SOLUTION: A research firm conducted a survey to determine the mean amount smokers spend on cigarette during a week. A sample of 49 smokers revealed that the sample mean is Br. 20 with standa

Question 1205253: A research firm conducted a survey to determine the mean amount smokers spend on cigarette during a week. A sample of 49 smokers revealed that the sample mean is Br. 20 with standard deviation of Br. 5. Construct 95% confidence interval for the mean amount spent.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

At 95% confidence, the z critical value is roughly z = 1.96 which is something to memorize or have handy on a reference sheet somewhere.

n = 49 = sample size
xbar = 20 = sample mean
s = 5 = sample standard deviation

E = margin of error for the mean
E = z*s/sqrt(n)
E = 1.96*5/sqrt(49)
E = 1.4

L = lower boundary of the confidence interval
L = xbar - E
L = 20 - 1.4
L = 18.6

U = upper boundary of the confidence interval
U = xbar + E
U = 20 + 1.4
U = 21.4

The 95% confidence interval is approximately 18.6 < mu < 21.4
It's of the format L < mu < U.

We can condense that to the notation (18.6, 21.4)

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