Question: A researcher is interested in whether English proficiency of children from an immigrant family is different from that of children that are citizens. The level of English proficiency was measured by a scale measuring verbal ability. The mean English proficiency of 100 children of immigrants was 300. You know that the mean English proficiency of children in the general population at the same age is 302 and standard deviation is 10. Conduct an appropriate statistical testing and report the conclusion.
a) Provide non-directional null and alternative hypotheses
b) Perform a z-test using alpha = .05. Report the calculated z-statistics and p-value
c)Calculate the 95% confidence interval.
d) Make a decision based on alpha and p-value.
e) Interpret the result  Edit

Answer: a. Null hypothesis, H0: mu=300
and Alternative hypthesis, H1: mu ≠ 300

b. z= (302-300)/(10/sqrt(100))=2
critical z at alpha=0.05 is 1.96

hence calculated z(2) >critical z(1.96)


p-value=P(|z|>2)=1-2*P(0 c. 95% CI=(x bar-10/sqrt(100)*1.96, x bar+10/sqrt(100)*1.96 )=(302-1.96, 302+1.96)=(300.04, 303.96)
d. hence p-value(0.5228 ) < alpha

Reject null .
e. English proficiency of children from an immigrant family is different from that of children that are citizens.
  Edit