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joja [24]
3 years ago
12

We draw a random sample of size 36 from a population with standard deviation 3.5. If the sample mean is 27, what is a 95% confid

ence interval for the population mean?
Mathematics
2 answers:
inessss [21]3 years ago
3 0

Answer:

The 95% confidence interval for the population mean is between 25.86 and 28.14.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.96\frac{3.5}{\sqrt{36}} = 1.14

The lower end of the interval is the sample mean subtracted by M. So it is 27 - 1.14 = 25.86

The upper end of the interval is the sample mean added to M. So it is 27 + 1.14 = 28.14

The 95% confidence interval for the population mean is between 25.86 and 28.14.

yaroslaw [1]3 years ago
3 0

Answer:

27-1.96\frac{3.5}{\sqrt{36}}=25.857  

27+1.96\frac{3.5}{\sqrt{36}}=28.143  

The 95% confidence interval would be given by (25.857;28.143). And we can conclude that the true mean with 95% is between (25.857;28.143)

Step-by-step explanation:

Data given

\bar X=27 represent the sample mean  

\mu population mean (variable of interest)  

\sigma=3.5 represent the population standard deviation  

n=36 represent the sample size  

95% confidence interval  

The confidence interval for the mean is given by the following formula:  

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)  

Since the Confidence is 0.95 or 96%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96  

Now we have everything in order to replace into formula (1):  

27-1.96\frac{3.5}{\sqrt{36}}=25.857  

27+1.96\frac{3.5}{\sqrt{36}}=28.143  

The 95% confidence interval would be given by (25.857;28.143). And we can conclude that the true mean with 95% is between (25.857;28.143)

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5 0
3 years ago
A party rental company has chairs and tables for rent. The total cost to rent 3 chairs and 2 tables is $18. The total cost to re
DerKrebs [107]
C= chair cost
t= table cost

Create two equations with the given information. Solve for one variable in equation one. Substitute that answer in equation two. Then you can solve for the needed information.
3c+2t=$18
5c+6t=$48

3c+2t=18
Subtract 2c from both sides
3c=18-2t
Divide both sides by 3
c=(18-2t)/3

Substitute the value for c in equation two:
5c+6t=$48
5((18-2t)/3)+6t=48
(90-10t)/3+6t=48
Multiply everything by 3 to eliminate fraction
(3)((90-10t)/3)+(3)(6t)=(3)(48)
90-10t+18t=144
90+8t=144
Subtract 90 from both sides
8t=54
Divide both sides by 8
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Substitute the t value to solve for c:
3c+2t=18
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3c+13.50=18
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Check:
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Hope this helps! :) If it does, please mark as brainliest.
7 0
3 years ago
If a(n)=24 which recursive formula could represent the sequence below? <br> ...,24,88,664,8408,...
sesenic [268]
24= 3*2^3  
88= 11*2^3
664= 83*2^3 -> 83=11+72 = 11 + 2^3*3^2
664=2^3 (11+2^3*3^2) = 88 +(2^3*2^3*3^2) = 88 +(24^2)

8408= 1051 * 2^3 -> 1051= 83+968 -> 968 = 2^3 * 11^2
8408= 2^3 (83+2^3*11^2) = 664 +(2^3*2^3*11^2) = 664 +(88^2)

So: 
a(n) = a(n-1) + a(n-2)^2

Lets check: 88+24^2= 664
664+88^2= 8408


3 1
3 years ago
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julsineya [31]
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The short answer is yes, 0.6 repeating is a rational number.

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Answer:

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Step-by-step explanation:

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