Question

A simple random sample of the sitting heights of 36 male students has a mean of 92.8cm. The population of males has sitting heights with a mean of 91.4 cm and a standard deviation of 3.6 cm. Using a 0.05 significance level to test the claim, the p-value of 0.0198 is found. What is the final conclusion about testing the claim that male students have a sitting height different from 91.4 cm. Could anyone show me how to do this?

Answer 1

SAMPLE STATISTICS: mean sitting height (x bar)=92.8cm, size (n)=36 male students.

POPULATION STATISTIC: mean sitting height (µ)=91.4cm, standard deviation (σ)=3.6cm.

CLAIM: Mean sitting height is different from 91.4cm.

COUNTERCLAIM: Mean sitting height is 91.4cm.

SIGNIFICANCE LEVEL: ɑ=0.05 (95% confidence level)

Null hypothesis, H₀: µ=91.4 (counterclaim)

Alternative hypothesis, H₁: µ≠91.4 (claim)

This is a 2-tail test, because “not equals” means less than (left tail) or greater than (right tail), so the significance level is split equally between the two tails giving us the required confidence interval of 95%.

TEST STATISTIC:

Z=(x bar-µ)/(σ/√n)=(92.8-91.4)/(3.6/√36)=1.4/0.6=2.333, corresponding to a P-value of 0.0098. The given test statistic is 0.0198.

ɑ/2=0.025 and sample P-value 0.0098<0.025, and 0.0198<0.025, which means that H₀ is rejected, and H₁ is accepted to be true. The conclusion is that the claim is true, that the mean sitting height of male students is different from 91.4cm.