Question

It is claimed that the average number of years of education of an employee is no...

It is claimed that the average number of years of education of an employee is no more than 12 years. Test this claim against sample results at a level of significance of 5% and draw a conclusion. How much weight of evidence does the sample data hold against the null hypothesis?

Population (n=93):

12,10,12,8,8,12,12,12,15,8,12,12,8,12,12,12,12,12,12,12,12,16,8,8,12,12,15,15,16,12,8,12,8,8,12,8,8,12,12,12,12,12,12,15,15,15,15,12,12,12,12,12,12,15,15,15,12,15,12,12,15,12,15,12,12,12,12,12,12,15,15,15,8,12,12,12,12,12,12,15,15,15,15,15,16,15,15,15,15,15,12,15,16

Math   Statistics-And-Probability   
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Answer #1

∑x = 1163

∑x² = 15023

n = 93

Mean , x̅ = Ʃx/n = 1163/93 = 12.5054

Standard deviation, s = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(15023-(1163)²/93)/(93-1)] = 2.2824

Null and Alternative hypothesis:

Ho : µ ≤ 12

H1 : µ > 12

Test statistic:

t = (x̅- µ)/(s/√n) = (12.5054 - 12)/(2.2824/√93) = 2.1354

df = n-1 = 92

p-value = T.DIST.RT(2.1354, 92) = 0.0177

Decision:

p-value < α, Reject the null hypothesis

Conclusion:

There is enough evidence to conclude that population mean is more than 12 at 0.05 significance level.

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