Question

A researcher believes that children who grow up as an only child develop vocabulary skills at a faster rate than children with siblings. To test this, a random sample of n = 10 four-year-old only children are tested on a standardized vocabulary test (mu (µ)=60, α=10). Can the researcher conclude that children from one-child families have greater vocabulary skills than the norm? Test at the .05 level. Use the data below to conduct the analysis.53,55, 59, 60, 62, 65, 68, 72, 73, 77Step

1: Develop Hypotheses:

a.Independent Variable = Scale: CategoricalQuantitative(1.5 pts)

b.Dependent Variable = Scale: CategoricalQuantitative (1.5 pts)

c.Circle: One-tailed Two-tailed (.05 pt)

d.Alternative hypothesis in sentence form (1 pt).

e.Null hypothesis in sentence form (1 pt).

f. Write the alternative and null hypotheses using correct notation (2 pts). H1:H0:

Step 2: Establish significance criteria (.05 pt)g.  =

Step 3: Calculate test statistic, effect size, confidence intervalh.zcalculated = Level of significance (p) =(1 pt)i.Decision: reject nullor fail to reject null(1 pt)j.Calculate effect size = (2 pts)k.Determine the 95% confidence interval: (1 pt)

Step 4: Draw conclusionl.Write your conclusion in sentence form including appropriate results notation (3 pts).

Answer 1

d. alternative hypothesis in sentence :

normal-child families have greater vocabulary than one-child families

e. null hypothesis in sentence :

one-child families have greater vocabulary skills than the norm

f.

H0 ( null hypothesis ) : one-child families have greater vocabulary skills than the norm

H0 : µ= 60

H1 ( alternative hypothesis) : normal-child families have greater vocabulary than one-child families

H1: µ 60

step2 : significance level of given hypothesis test is 0.05

step3 : test statistic :  

t= ( x̄ -µ ) / (s/√n)

mean= (sum of observations)/( number of observations)

x̄ = (53+55+ 59+ 60+ 62+ 65+ 68+ 72+ 73+ 77)/10

= 64.4

standard deviation (s) = 8.0027

t = (64.4-60)/ ( 8.0027/)

= 4.4/2.5307

=1.7234

confidence intervals =

  

t value at 95% level significance 2.262

= 64.4 2.262 * 2.5307

= 64.4 5.7244

confidence intervals ( 58.6756, 70.1244 )

step4 : t ( 1.7234) < t9 at 95% (2.262)

so null hypothesis H0 is accepted

i.e children from one-child families have greater vocabulary skills than the normal