Mathematics, 24.01.2020 21:31 vhife4901

Two random samples, a and b, were selected from the same population to estimate the population mean. for each sample, the mean, standard deviation, and margin of error for a 95 percent confidence interval for the population mean are shown in the table. mean standard deviation margin of error sample a 45 6.45 1.02sample b 43 7.84 0.72which of the following could explain why the margin of error of sample a is greater than the margin of error sample b? (a) the sample size of a is greater than the sample size of b. (b) the sample size of a is less than the sample size of b. (c) the sample size of a is equal to the sample size of b. (d) the mean of sample a is greater than the mean of sample b. (e) the standard deviation of sample a is less than the standard deviation of sample b.

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Answer from: breella0

$n_A = \frac{6.45^2}{(\frac{1.02}{1.96})^2}=153.61 \approx 154$

$n_B = \frac{7.84^2}{(\frac{0.72}{1.96})^2}=455.49 \approx 456$

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:

$\bar X_A= 45$ represent the sample mean for A

s_A= 6.45 represent the sample deviation for A

ME_A = 1.02 represent the margin of error for A

$\bar X_B= 43$ represent the sample mean for B

s_B= 7.84 represent the sample deviation for B

ME_B= 0.72 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 $\hat \sigma = \hat s$ and we can use the following formula for the margin of error of the sample mean the following formula:

$ME= z_{\alpha/2} \frac{\hat s}{\sqrt{n}}$

For this case the value of the significance is given by $\alpha =1-0.95 =0.05$ and the value for $\alpha/2 =0.025$ , so then the value for $z_{\alpha/2}$ 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 $z_{\alpha/2}=\pm 1.96$ .

So then since we have the value for z if we solve for n from the margin of error formula we got:

$n = \frac{\hat s^2}{(\frac{ME}{z})^2}$

And for the case A we can find the sample size and we got:

$n_A = \frac{6.45^2}{(\frac{1.02}{1.96})^2}=153.61 \approx 154$

And for the case B we can find the sample size and we got:

$n_B = \frac{7.84^2}{(\frac{0.72}{1.96})^2}=455.49 \approx 456$

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.

Answer from: Quest

there has been genes found in the x chromosome of males that have been linked to [homo/hetero]sexuality. in addition to the reign of the x chromosome, a section of chromosome 8 has also been link to, "alot of ppl are so " these facts were proven true after a 2014 study of a gay male and his strait brother. according to newscientist.com,"chromosome 14 and is mainly active in the thyroid, but also the brain. called tshr, it makes a type of receptor protein that recognises and binds to a hormone that stimulates the thyroid. in this way, the gene plays an important role in controlling thyroid function. the fact that tshr seems to be involved in sexual orientation fits with evidence that thyroid function seems to be linked to sexuality."

in conclusion, "ppl are so gay" due to their genetics.

Answer from: Quest

step-by-step explanation:

51/3=17

thus 17 is your answer.

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Two random samples, a and b, were selected from the same population to estimate the population mean....