Question

1. An investigator uses children’s WISC IQ scores as part of her study of how children respond to different teaching methods. She obtains a random sample of four children from a 5^th grade class and obtains their WISC IQ scores. Their scores are shown below. Her null hypothesis is that her sample does not differ from the general population of 5^th graders, whose IQ scores are assumed to have a m of 100 and a s of 15. She decides to use the known m and s, adopts an a of .05, and uses a non-directional (i.e., 2-sided) alternative hypothesis.

IQ scores: 112, 110, 123, 115

Perform the null hypothesis significance test that is appropriate for this situation. Show your work; Summarize your results and briefly state what they mean.

Answer 1

set up hypothesis :

Null hypothesis :

WHERE IS POPULATION MEAN .

Alternative hypothesis :

clearly sample size is 4 which is less than 30 , we should have to use t-test ( Student's t-distribution )

the observations are given as 112,110,123,115

Sample mean =

:here is output of R-software

> x=c(112, 110, 123, 115)
> x
[1] 112 110 123 115
> t.test(x,mu=100,conf.level=0.95)

One Sample t-test

data: x
t = 5.2489, df = 3, p-value = 0.01347
alternative hypothesis: true mean is not equal to 100
95 percent confidence interval:
105.9054 124.0946
sample estimates:
mean of x
115

Decision rule :

THEN WE REJECT NULL HYPOTHESIS AT \alpha =0.05

HERE 0.01347<0.05

DECISION : We can reject null hypothesis i.e accept alternative hypothesis

Conclusion : sample differ from the general population