Respuesta :
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 [tex]\mu_1[/tex] = population mean score for children born to cocaine users.
[tex]\mu_2[/tex] = 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. = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ~ [tex]t__n_1_+_n_2_-_2[/tex]
where, [tex]\bar X_1[/tex] = sample mean score of children born to cocaine users = 7.3
[tex]\bar X_2[/tex] = sample mean score of children not exposed to cocaine = 8.2
[tex]s_1[/tex] = sample standard deviation for children born to cocaine users = 3
[tex]s_2[/tex] = sample standard deviation for children not exposed to cocaine = 3
[tex]n_1[/tex] = sample of children born to cocaine users = 190
[tex]n_2[/tex] = sample of children not exposed to cocaine = 186
Also, [tex]s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(190-1)\times 3^{2}+(186-1)\times 3^{2} }{190+186-2} }[/tex] = 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.
In the null hypothesis, a test always forecasts no effect, while the alternate theory states the research expectation impact, and calculation as follows:
Null and alternative hypothesis:
Calculating the pooled estimator of [tex]\sigma^2[/tex], denoted by [tex]S^2_p[/tex], is defined by
[tex]\to \bold{S^2_p= \frac{(n_1 - 1) S^2_1+ (n_2 - 1)S^2_2}{n_1 + n_2 - 2}}[/tex]
Null hypothesis:
[tex]\to H_0 : \mu_1 - \mu_2 = \Delta_0\\[/tex]
Test statistic:
[tex]\to T_0=\frac{\bar{X_1}- \bar{X_2} -\Delta_0}{S_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \\\\[/tex]
Alternative Hypothesis:
[tex]H_1 : \mu_1 -\mu_2 \neq \Delta_0\\\\ H_1 : \mu_1 -\mu_2 > \Delta_0\\\\H_1 : \mu_1 -\mu_2 < \Delta_0\\\\[/tex]
Rejection Criterion
[tex]t_0 > t_{\frac{\alpha}{2} , n_1+n_2 -2}\ \ \ or\ \ \ t_0 < - t_{\frac{\alpha}{2} , n_1+n_2 -2} \\\\t_o > t_{\alpha , n_1+n_2 -2} \\\\t_o > -t_{\alpha , n_1+n_2 -2}[/tex]
Given value:
[tex]\to S_p=9\\\\\to \Delta_0=0\\\\\to t_0=-\frac{0.9}{3(\sqrt{(\frac{1}{190}+\frac{1}{186})})}=-2.9\\\\\to t_{0.05,374}=1.645\\\\[/tex]
here
[tex]\to t_0 < -t_{0.05,374}[/tex]
hence rejecting the [tex]H_0[/tex]
Since there should be enough evidence that prenatal cocaine exposure is linked to inferior item assembly scores in 4-year-olds.
Find out more about the alternative hypothesis here:
brainly.com/question/18831983
