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# 305, ex 5 a. \$20 n=49 Xbar = 20 20 +/- 1.96(5)/sqrt(49) pop. dev = 5 95% CI on mean spending 20 1.4 95%CL Z = 1.96 LL 18.6 Error= 1.40 UL 21.

P. 305, ex 5 a. \$20
n=49
Xbar = 20 20 +/- 1.96(5)/sqrt(49)
pop. Std. dev = 5
95% CI on mean spending 20 1.4
95%CL Z = 1.96
LL 18.6
Error= 1.40 UL 21.4

6. Refer to the previous exercise. Suppose that 64 smokers (instead of 49) were sampled.
Assume the sample mean remained the same.

n=64
xbar=20
pop. Std. dev =5
95% Cl on mean spending
95%CL Z = 1.96

a. What is the 95 percent confidence interval estimate of µ ?
b. Explain why this confidence interval is narrower than the one determined in the
previous exercise.

Dr. Patton is a professor of English. Recently she counted the number of misspelled

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