Question

Suppose the mean height of women age 20 years or older in a certain country is 62.9 inches. One hundred randomly selected women in a certain city had a mean height of 63.8 inches. At the 5% significance level, do the data provide sufficient evidence to conclude that the mean height of women in the city differs from the national mean? Assume that the population standard deviation of the heights of women in the city is 3.9 inches. The standard deviation is of population. What is the parameter of interest? What is the underlying distribution?

Answer 1

Parameter of interest : population mean

We will use z distribution.

x̅ = 63.8, σ = 3.9, n = 100

Null and Alternative hypothesis:

Ho : µ = 62.9

H1 : µ ≠ 62.9

Critical value :

Two tailed critical value, z crit = ABS(NORM.S.INV(0.05/2)) = 1.960

Reject Ho if z < -1.96 or if z > 1.96

Test statistic:

z = (x̅- µ)/(σ/√n) = (63.8 - 62.9)/(3.9/√100) = 2.3077

p-value :

p-value = 2*(1-NORM.S.DIST(ABS(2.3077), 1)) = 0.0210

Decision:

p-value < α, Reject the null hypothesis

Conclusion:

There is enough evidence to conclude that the mean height of women in the city differs from the national mean at 0.05 significance level.