We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

$\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}$

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

$CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack$

Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

$CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack$

Where (from tables):

$Z_{0.99}=2.33$

Finally, the interval at 98% confidence level is:

$CI(\mu)=\lbrack28.94,31.06\rbrack$

4 0

1 year ago

A polynomial P(x) and a divisor d(x) are given. Use long division to find a quotient Q(x) and the remainder R(x). Express P(x) i

Vilka [71]

It's a bit hard to explain but I hope this helps

5 0

4 years ago

PLEASE HELP ASAP I NEED THIS!

Misha Larkins [42]

Answer:

it would be A

Step-by-step explanation:

Because Coaster A is TWICE as large as coaster B 15+15 is 30

3 0

3 years ago

Express the following as a pecentage:16 out of 40

Irina-Kira [14]

the answer is 40 percent

6 0

3 years ago