Answer :
Answer:
[tex] n_A = \frac{6.45^2}{(\frac{1.02}{1.96})^2}=153.61 \approx 154[/tex]
[tex] n_B = \frac{7.84^2}{(\frac{0.72}{1.96})^2}=455.49 \approx 456[/tex]
For this case as we can see we have a larger sample size for sample B, so then the best option for this case would be:
(B) The sample size of A is less than the sample size of B.
Step-by-step explanation:
For this case we have the following data given:
[tex] \bar X_A= 45[/tex] represent the sample mean for A
[tex] s_A= 6.45[/tex] represent the sample deviation for A
[tex] ME_A = 1.02[/tex] represent the margin of error for A
[tex] \bar X_B= 43[/tex] represent the sample mean for B
[tex] s_B= 7.84[/tex] represent the sample deviation for B
[tex] ME_B= 0.72[/tex] represent the margin of error for B
And for this case we are assuming that we have the same confidence level of 95%
For this case we an use the fact that the sample deviation is an unbiased estimator for the population deviation [tex]\hat \sigma = \hat s[/tex] and we can use the following formula for the margin of error of the sample mean the following formula:
[tex] ME= z_{\alpha/2} \frac{\hat s}{\sqrt{n}}[/tex]
For this case the value of the significance is given by [tex] \alpha =1-0.95 =0.05[/tex] and the value for [tex]\alpha/2 =0.025[/tex] , so then the value for [tex] z_{\alpha/2}[/tex] represent a quantile of the normal standard distribution that accumulates 0.025 of the area on each tail of the normal standard distribution and for this case is [tex] z_{\alpha/2}=\pm 1.96[/tex].
So then since we have the value for z if we solve for n from the margin of error formula we got:
[tex] n = \frac{\hat s^2}{(\frac{ME}{z})^2}[/tex]
And for the case A we can find the sample size and we got:
[tex] n_A = \frac{6.45^2}{(\frac{1.02}{1.96})^2}=153.61 \approx 154[/tex]
And for the case B we can find the sample size and we got:
[tex] n_B = \frac{7.84^2}{(\frac{0.72}{1.96})^2}=455.49 \approx 456[/tex]
For this case as we can see we have a larger sample size for sample B, so then the best option for this case would be:
(B) The sample size of A is less than the sample size of B.
The sample size of B is larger than the sample size of A and this can be determined by using the formula of margin of error.
Given :
- Two random samples, A and B, were selected from the same population to estimate the population mean.
- 95 percent confidence interval.
- The sample mean for A = 45
- The sample deviation for A = 6.45
- The margin of error for A = 1.02
- The sample mean for B = 43
- The sample deviation for B = 7.84
- The margin of error for B = 0.72
To determine the sample size for both cases A and B, the formula of Margin of Error can be used:
[tex]\rm ME =z_{\frac{\alpha }{2}} \dfrac{\hat{s}}{\sqrt{n} }[/tex]
[tex]\rm n =\left(\dfrac{\hat{s}}{\dfrac{ME}{z}}\right)^2[/tex]
Now, for case A:
[tex]\rm n_A =\left(\dfrac{6.45}{\dfrac{1.02}{1.96}}\right)^2[/tex]
[tex]\rm n_A\approx 154[/tex]
Now, for case B:
[tex]\rm n_B =\left(\dfrac{7.84}{\dfrac{0.72}{1.96}}\right)^2[/tex]
[tex]\rm n_B \approx 456[/tex]
So, the sample size of B is larger than the sample size of A.
Therefore, the correct option is B) The sample size of A is less than the sample size of B.
For more information, refer to the link given below:
https://brainly.com/question/21586810