Question

A study was conducted to assess the effects that occur when children are exposed to cocaine before birth. Children were tested at age 4 for object assembly skill, which was described as a task requiring visual spatial skills related to mathematical competence. The 190 children born to cocaine users had a mean of 7.3 and a standard deviation of 3.0 The 186 children not exposed to cocaine had a mean score of 8.2 with a standard deviation of 3.0 Use a 0.05 significance level to test the claim that prenatal cocaine exposure is associated with lower scores of four year old children on the test of object assembly.
What are null and alternative hypothesis? What is test statistics?

Answer 1

Answer:

We conclude that prenatal cocaine exposure is associated with lower scores of four-year-old children on the test of object assembly.

Step-by-step explanation:

We are given that the 190 children born to cocaine users had a mean of 7.3 and a standard deviation of 3.0 The 186 children not exposed to cocaine had a mean score of 8.2 with a standard deviation of 3.0.

Let $\mu_1$ = population mean score for children born to cocaine users.

$\mu_2$ = population mean score for children not exposed to cocaine.

So, Null Hypothesis, : = 490      {means that the prenatal cocaine exposure is not associated with lower scores of four-year-old children on the test of object assembly}

Alternate Hypothesis, : 490     {means that the prenatal cocaine exposure is associated with lower scores of four-year-old children on the test of object assembly}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about population standard deviations;

                               T.S.  =  $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$  ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean score of children born to cocaine users = 7.3

$\bar X_2$ = sample mean score of children not exposed to cocaine = 8.2

$s_1$ = sample standard deviation for children born to cocaine users = 3

$s_2$ = sample standard deviation for children not exposed to cocaine = 3

$n_1$ = sample of children born to cocaine users = 190

$n_2$ = sample of children not exposed to cocaine = 186

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$  = $\sqrt{\frac{(190-1)\times 3^{2}+(186-1)\times 3^{2} }{190+186-2} }$ = 3

So, the test statistics =   ~

                                     =  -2.908

The value of t-test statistics is -2.908.

Now, at a 0.05 level of significance, the t table gives a critical value of -1.645 at 374 degrees of freedom for the left-tailed test.

Since the value of our test statistics is less than the critical value of t as -2.908 < -1.645, so we have sufficient evidence to reject our null hypothesis as the test statistics will fall in the rejection region.

Therefore, we conclude that prenatal cocaine exposure is associated with lower scores of four-year-old children on the test of object assembly.