Question

Question 12 8 pts A study of 80 mice showed that their average weight was 6.8 ounces. The standard deviation of the population is 0.9 ounces. Which of the following is the 99% confidence interval for the mean weight per mouse? (5.4408, 8.7502) (6.0408,7.9502) (6.5408, 7.0592) (5.9408,8.1502) Question 13 8 pts 57 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 13,5 pounds; the population standard deviation for all packages is known to be 3.1 pounds. Based on this data, what is the 95 percent confidence interval for the true mean weight of all packages received by the parcel service? (12.695, 16,305) (10.695, 16.305) (10.695, 12.305) (12.695, 14.305)

Answer 1

Question 12

The 99% confidence interval for the mean weight per mouse is given by,

CI = (T + za/2 x 이트 12

Where, $\bar{x}=6.8,\sigma =0.9,n=80$

$1-\alpha =0.99\Rightarrow \alpha =0.01\Rightarrow \alpha /2=0.005$

We will get the value of $z_{\alpha /2}$ from the last row of t-table.

From t-table,  

2a/= 70.005 2.576

$\therefore CI=(6.8\pm 2.576\times \frac{0.9}{\sqrt{80}})$

$=(6.8\pm 2.576\times 0.1006)=(6.8\pm 0.2592)$

$=(6.5408,7.0592)$

The 99% confidence interval for the mean weight per mouse is,

(6.5408,7.0592)

Question 13

The 95% confidence interval for the true mean weight of all packages received by the parcel service is given by,

CI = (T + za/2 x 이트 12

where, $\bar{x}=13.5,\sigma =3.1,n=57$

$1-\alpha =0.95\Rightarrow \alpha =0.05\Rightarrow \alpha /2=0.025$

We will get the value of $z_{\alpha /2}$ from the last row of t-table.

From t-table,  $z_{\alpha /2}=z_{0.025}=1.96$

$\therefore CI=(13.5\pm 1.96\times \frac{3.1}{\sqrt{57}})$

$=(13.5\pm 1.96\times 0.4106)=(13.5\pm 0.8048)$

$=(12.695,14.305)$

The 95% confidence interval for the true mean weight of all packages received by the parcel service is,

(12.695,14.305)

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