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Twenty children with mild autism were selected from schools throughout the state

ID: 3385515 • Letter: T

Question

Twenty children with mild autism were selected from schools throughout the state of Illinois for assessment on a social skills inventory. Typical seven-year-olds have a mean score of 275 on this inventory, but autistic children will probably score lower. Given the following social inventory scores for the 20 autistic children, carry out the appropriate statistical analysis.

            180      265      225      250      205

            255      205      235      215      260     

            215      185      210      260      200

            195      225      195      225      195

a.) State the null and alternative hypotheses.

b.) Calculate the appropriate standard error of the mean.

c.) Calculate the appropriate statistic (z or t).

d.) Is this a one- or two-tailed test?

e.) What is (are) the critical value (s) given a = 0.05?

f.) Should the research retain or reject Ho?

g.) Indicate the 95% confidence intervals and the effect size (d).

Explanation / Answer

let X be the random variable denoting the score for assessment on a social skills inventory for autistic children.

so X~N(mu,sigma2)

now here sigma is unknown.

we have a random sample of size n=20 from X and we want to test whether mu is less thab 275 or not

a) null hypothesis H0: mu=275    vs alternative hypothesis H1:mu<275

b) the standard error of the mean is sample SD/sqrt(sample size=20)=26.5072/sqrt(20)=4.472136

c) as here the population standard deviation is unknown here the the test statistic should be a t statistic with degrees of freedom n-1

the test statistic is T=(Xbar-275)/(standard error of the mean) which under H0 follows a t distribution with degrees of freedom n-1=20-1=19           [xbar is the sample mean]

d) as here H1:mu<275   hence the test is a one tailed test.

we reject H0 iff the observed value of T is less than the critical value -ta,19. where ta,19 denotes the upper 100*(1-a)% point of a t distribution with degrees of freedom n-1=20-1=19

e) so the critical value at a=0.05 is -1.72913

f) now Xbar=sample mean=220

so observed value of T is (220-275)/4.472136=-12.29837

which is less than -1.72913

hence the test reject H0 and concludes that the autistic children score lower in assessment on a social skills inventory

g) the 95% confidence interval is

[Xbar-standard error*t0.05,19,Xbar+standard error*t0.05,19]=[220-4.472136*1.72913,220+4.472136*1.72913]=[212.2671,227.7329]

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