183k views
4 votes
A researcher was interested in comparing the heights of women in two different countries. Independent simple random samples of 9 women from country A and 9 women from country B collected data for their heights (in inches). The following 90% confidence interval was obtained for the difference between the mean height of women in country A and the mean height of women in country B. -4.34 in. < μA - μB < -0.03 in What does the confidence interval suggest about the population means?

User Wonay
by
4.9k points

1 Answer

3 votes

Answer:

Null hypothesis:
\mu_(A)-\mu_(B)= 0

Alternative hypothesis:
\mu_(A)-\mu_(B) \\eq 0

And for this case we can use the confidence interval given by:


\bar X_A -\bar X_b \pm t_(\alpha/2) \sqrt{(s^2_A)/(n_A) +(s^2_B)/(n_B)}

And after calculate the 90% confidence interval we got:


-4.34 < \mu_A -\mu_B < -0.03

So then we see that all the values are negative so then we can conclude that the two means are different and on this case the mean for A seems to be lower than the mean for B at 10% of significance.

Explanation:

Data given and notation


\bar X_(A) represent the mean for the sample A


\bar X_(B) represent the mean for the sample B


s_(A) represent the sample standard deviation for the sample A


s_(B) represent the sample standard deviation for the sample B


n_(A) sample size selected A


n_(B) sample size selected B


\alpha=0.1 represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)


p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the two means are equal., the system of hypothesis would be:

Null hypothesis:
\mu_(A)-\mu_(B)= 0

Alternative hypothesis:
\mu_(A)-\mu_(B) \\eq 0

And for this case we can use the confidence interval given by:


\bar X_A -\bar X_b \pm t_(\alpha/2) \sqrt{(s^2_A)/(n_A) +(s^2_B)/(n_B)}

And after calculate the 90% confidence interval we got:


-4.34 < \mu_A -\mu_B < -0.03

So then we see that all the values are negative so then we can conclude that the two means are different and on this case the mean for A seems to be lower than the mean for B at 10% of significance.