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A random sample of 10 single mothers was drawn from a Obstetrics Clinic. Their ages are as follows: 22 17 27 20 23 19 24 18 19 24 We want to determine at the 5% significance level that the population mean is not equal to 20. What is the rejection region?

User Zorawar
by
5.1k points

1 Answer

0 votes

Answer:

0.09

Explanation:

Let x = ages of mother

x : 22 17 27 20 23 19 24 18 19 24

N = 10

Mean = ∑x/N = 218/10 = 21.8

Difference in mean = 21.8 - 20 = 1.8

If significance level = 5% or 0.05

∴ Rejection region = 1.8 X 0.05 = 0.09

User Valeri
by
4.8k points
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