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

Using a video camera positioned on a table, Kat records her hanging flower basket to find out what is eating the blooms. The cam

Serga [27]

Answer: 75.5

I took the test and got it right.

8 0

3 years ago

The area of the shaded segment is 100cm^2. Calculate the value of r.

Reil [10]

Hello,

The formula for finding the area of a circular region is: $A= \frac{ \alpha *r^{2} }{2}$

then:
$A_{1} = \frac{80*r^{2} }{2}$

With the two radius it is formed an isosceles triangle, so, we must obtain its area, but first we obtain the height and the base.

$cos(40)= \frac{h}{r} \\ \\ h= r*cos(40)\\ \\ \\ sen(40)= \frac{b}{r} \\ \\ b=r*sen(40)$

Now we can find its area:
$A_{2}=2* \frac{b*h}{2} \\ \\ A_{2}= [r*sen(40)][r*cos(40)]\\ \\A_{2}= r^{2}*sen(40)*cos(40)$

The subtraction of the two areas is 100cm^2, then:

$A_{1}-A_{2}=100cm^{2} \\ (40*r^{2})-(r^{2}*sen(40)*cos(40) )=100cm^{2} \\ 39.51r^{2}=100cm^{2} \\ r^{2}=2.53cm^{2} \\ r=1.59cm$

Answer: r= 1.59cm

7 0

3 years ago

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2.4 x 24.12 need help I’m stuck help

Degger [83]

Answer: 57.888

Step-by-step explanation: you have to use long multiplication to evaluate

4 0

2 years ago

Read 2 more answers