Answer to Question #255108 in Statistics and Probability for Lexie

Question

Use the traditional method in testing hypothesis. Developmental psychologists at a prominent California university conducted a longitudinal study investigating the effect of high levels of curiosity in early childhood on intelligence. The local population of 3-year-olds was screened via a test battery assessing curiosity. Twelve of the 3-year-olds scoring in the upper 90% of this variable were given an IQ test at age 3 and again at age 11. The following IQ scores were obtained. Student Number: 1 2 3 4 5 6 7 8 9 10 11 12 IQ (Age of 3) : 100 105 125 140 108 122 117 112 135 128 104 98 IQ (Age of 11) : 114 116 139 151 106 119 131 136 148 139 122 113 Using ∝= 0.01, two-tailed test, analyse the data and then interpret the result.

Answer 1

In this question, we have a case where our sample is dependent and we should first consider using matched/paired t-test for our analysis.

To perform this test, we begin by finding the difference between pairs of observation for each child. We shall use the notation $d,$ to represent the differences and use it to make a decision.

The hypotheses to be tested are,

$H_0:\mu_d=0$

$Against$

$H_1:\mu_d\not=0$, where $d$ is the difference between each pair of observations.

To test these hypotheses, we first determine the mean of the differences$(\bar{d})$ given by,

$\bar{d}=(\sum(d))/n$, where $n=12.$

Now,

$\sum(d)=-140$

Therefore,

$\bar{d}=-140/12=-11.66666667$

Variance of the differences is given by,

$V(d)=\sum(d-\bar{d})^2/(n-1)$

$V(d)=624.666666667/11=56.78788(5\space d\space p)$.

The standard deviation $SD(d)$ is given by,

$SD(d)=\sqrt{V(d)}=\sqrt{56.78788}=7.535773271$.

The test statistic is given by,

$t^*=\bar{d}/(SD(d)/\sqrt n)$

$t^*=-11.66666667/(7.535773271/\sqrt {12})$

$t^*=-11.66666667/2.175390363=-5.363022133.$

$t^*$ is compared with the student's t distribution with $(n-1)=12-1=11$ degrees of freedom at $\alpha=0.01$.

The table value $t_{\alpha/2,11}=t_{0.01/2,11}=t_{0.005,11}=3.106$ and reject the null hypothesis if $|t^*|\gt t_{0.005,11}.$

Since $|t|=5.363\gt t_{0.005,11}=3.106,$ we reject the null hypothesis and conclude that sufficient evidence exists to show that high levels of curiosity in childhood affect IQ. They seem to increase the IQ.