Question

Suppose we have two independent, normally distributed, populations and we draw a random sample from each population. For population A, we drew a sample of 13 and found the mean to be 17 with a standard deviation of 4. For population B, we drew a sample of 12 with a mean of 19 and a standard deviation of 3. We want to determine if the treatments result in the same effect. Show all work.

Answer 1

To determine if the treatments result in the same effect, we can use the two-sample t-test to compare the means of the two populations.

Step-by-step explanation:

To perform a hypothesis test in this scenario, we can use the two-sample t-test. This test is used to compare the means of two independent populations, assuming that the populations are normally distributed. The random variable in this case would be the difference between the means of the two samples.

In order to perform the hypothesis test, we need to define the null and alternative hypotheses. In this case, the null hypothesis would be that the treatments result in the same effect, meaning that there is no significant difference between the two population means. The alternative hypothesis would be that the treatments result in different effects, indicating a significant difference between the two population means. We can then calculate the test statistic and compare it to the critical value to make a decision.

In summary, to determine if the treatments result in the same effect, we can use the two-sample t-test to compare the means of the two populations. We need to define the null and alternative hypotheses, calculate the test statistic, and compare it to the critical value to make a decision.

Answer 2

Since the calculated value of z= -1.421 does not lie in the critical region z >

- 1.96 the null hypothesis is accepted and it is concluded that the treatments result in the same effect.

Step-by-step explanation:

1) Let the null and alternate hypothesis be

H0: μa − μb = 0 the treatments result in the same effect

against the claim

Ha: μa − μb ≠ 0 the treatments do not result in the same effect

2) The significance level is set at 0.05

3) The critical region is z = ± 1.96

4) The test statistic

Z= x1`-x2`/ sqrt [( s1²/n1+ s2²/n2)]

Here

x1`= 17 s1= 4 n1= 13

x2`= 19 s2= 3 n2= 12

5) Calculations

Z= x1`-x2`/ sqrt [( s1²/n1+ s2²/n2)]

z= 17-19/sqrt [( 16/13+ 9/12)]

z= -2/1.40739

z= -1.421

6) Conclusion

Since the calculated value of z= -1.421 does not lie in the critical region z >

- 1.96 the null hypothesis is accepted and it is concluded that the treatments result in the same effect.