Question

QUESTION 2 [10 pts total] The National Toxicology Program evaluates the toxicity of chemicals found in manufacturing in consumer products, or in the environment after disposal. Here are some of the results from a study of the toxicity of black newspaper ink in 7-week old female rats. One group of rats was exposed to ink dermally twice a week for 13 weeks while the control group of rats was left untreated. The following table contains body weights at the beginning and end of the study in grams.

control		exposed
week 0	week 13	week 0	week13
111.2	191.6	107.3	187
105.4	191.2	116.7	189.5
110.8	210.7	112.2	179.3
105.6	185.2	103.4	172.2
106.1	195	113.2	178.7
104.4	188.3	110.6	180.9
114	188.4	110.6	188.3
115.1	195.6	100.5	188.9
109.2	204.6	106.3	183.1
111.3	195.7	112.5	184.5

2A) [5 pts] Verify the two experimental groups do not have significantly different weights at the beginning of the study. State the null hypothesis and alternative hypothesis, then either make a convincing argument by presenting a confidence interval or with a p-value.

2B) [5 pts] Is there good evidence that ink application impairs growth in female rats? State the null hypothesis and alternative hypothesis then use a Two-Sample t-test Assuming Equal Variances to make a convincing argument.

- we are aloud to use excel

Answer 1

2A) Excel output of the test:

t-Test: Two-Sample Assuming Equal Variances
	Control	Exposed
	week 0	week 0
Mean	109.31	109.33
Variance	14.30544444	24.13788889
Observations	10	10
Pooled Variance	19.22166667
Hypothesized Mean Difference	0
df	18
t Stat	-0.010200453
P(T<=t) one-tail	0.495986781
t Critical one-tail	1.734063592
P(T<=t) two-tail	0.991973563
t Critical two-tail	2.100922037

Let us denote population mean weight of female rats at week 0 of two groups by .

We have to test whether mean weight is significantly different in two groups in the beginning or not.

Null hypothesis:

Alternative hypothesis:

Test statistic:

Find S_p under the assumptions of equal variances.

Therefore, we get test statistic as:

p-value for two tailed test:

p-value = 0.9921

Since, we fail to reject null hypothesis.

There is sufficient evidence to support the claim that two experimental groups do not have significantly different weights.

2B) Excel output:

t-Test: Two-Sample Assuming Equal Variances
	Control	Exposed
	week 13	week13
Mean	194.63	183.24
Variance	60.95788889	30.67377778
Observations	10	10
Pooled Variance	45.81583333
Hypothesized Mean Difference	0
df	18
t Stat	3.76271159
P(T<=t) one-tail	0.000712504
t Critical one-tail	1.734063592
P(T<=t) two-tail	0.001425008
t Critical two-tail	2.100922037

Let us denote population mean weight of female rats at week 13 of two groups by .

We have to test whether mean weight in exposed group is increased.

Null hypothesis:

Alternative hypothesis:

Test statistic:

Find S_p under the assumptions of equal variances.

Therefore, we get test statistic as:

p-value for one tailed test:

p-value = 0.00071

Since, we reject null hypothesis.

There is sufficient evidence to support the claim that ink application impairs growth in female rats.