Question

Please Help!!! :(
Carlos is analyzing the results of an experiment on two groups of mice in which he compared the difference in their weight gain after feeding each group a different type of food for one month. The difference in the mean weight gain is 0.6 grams. The standard deviation of the difference in sample means is 0.305.
The 68% confidence interval for the population mean difference is ±???
The 95% confidence interval for the population mean difference is ±???

Answer 1

so do I first have to look for the z score?
z=-0.51

Answer 2

Confidence interval for population mean difference
= difference in sample mean weight ± z score * standard deviation

Answer 3

we know that , by the empirical rule, within one standard deviation is 68% of the data
and 95% of the data is within 2 standard deviations. Therefore the z score for 68% two tailed is z=1, and z score for 95% two tailed is z = 2.

68% confidence interval: .6 ± 1* 0.305
95% confidence interval: .6 ± 2* 0.305

Answer 4

so would I write it like this?
The 68% confidence interval for the population mean difference is ± 0.905
The 95% confidence interval for the population mean difference is ± 1.21

Answer 5

not quite

Answer 6

.6 ± 1* 0.305 = ( .6 - 0.305 , .6 + 0.305)

Answer 7

for this one it isn't correct.
The 95% confidence interval for the population mean difference is ± 1.21
it is one of these
0.160 , 0.305, 0.610, 0.915

Answer 8

.6 ± 1* 0.305 = ( .6 - 0.305 , .6 + 0.305) = ( .295, .905)
.6 ± 2* 0.305 = ( -.01, 1.21)

Answer 9

They would both be 0.305

Answer 10

can you take a screen shot of this?

Answer 11

I think it should say .6 ± .305, but for some reason they left out the .6

Answer 12

Im confused as to why they would be the same .305

Answer 13

because it can be greater or less than .305