N a triangle, the measure of the first angle is twicetwice the measure of the second angle. the measure of the third angle is 6
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Answer:
a) Sample Mean = 79.4
Sample standard deviation = 30.62
b) 90% Confidence interval: (67.56 ,91.24)
Step-by-step explanation:
We are given the following in the question:
Prices for sleeping bags has an approximately normal distribution.
We are given the following sample:
35, 85, 105, 40, 100, 50, 30, 23, 100, 110, 105, 95, 105, 60, 110, 120, 95, 90, 60, 70
a)
Formula:
where are data points,
is the mean and n is the number of observations.
Sum of squares of differences = 1971.36 + 31.36 + 655.36 + 1552.36 + 424.36 + 864.36 + 2440.36 + 3180.96 + 424.36 + 936.36 + 655.36 + 243.36 + 655.36 + 376.36 + 936.36 + 1648.36 + 243.36 + 112.36 + 376.36 + 88.36 = 17816.8
b) 90% Confidence interval:
Putting the values, we get,
Answer:
A. [12.92;14.97]lb
B. [13.25;16.26]%
C. check explanation
Step-by-step explanation:
Hello!
To properly resolve any statistical problem you need to first identify your study variables and summarize the given data.
A.
Study variable:
X: "backpack weight of a sixth-grader" (lb)
sample n= 131
sample mean: x[bar]= 13.83 lb
sample standard deviation: S= 5.05 lb
Now you need to calculate a CI for the population mean of the backpack weight of sixth graders (μ) with a confidence level of 99%
Since you need to estimate the population mean, and you have a sample large enough (n≥30), although you don't know the distribution of this population, you can approximate the distribution of the sample mean to normal (applying CLT) and use the statistic Z= (x[bar] - μ)/(S/√n) ≈ N(0;1)
The formula for the CI is
x[bar] ± * (S/√n)
The confidence level to use is 1-α = 0.99
α = 0.01 ⇒ α/2 = 0.005
⇒
= -2.58
Now you calculate the interval
13.83 ± 2.58 * 5.05/√131 ⇒ [12.92;14.97]lb
So with a confidence level of 99%, you'll expect that the real mean for the backpack weight of sixth graders is contained by the interval [12.92;14.97]lb
B.
It seems that for this item the information was measured from the same sample taken in item A.
Study variable
X: "Backpack weight as percentage of body weight" (lb)
Assuming is the same sample: n= 131
It was calculated the 95% CI [13.62;15.89]lb
And you need to calculate a 99% CI
For this item, you use the same statistic as before
x[bar] ± * (S/√n)
To calculate the asked interval you need to know the confidence level, the sample mean, sample standard deviation and sample size.
To know the sample mean and standard deviation for the backpack weight as percentage of body weight, you'll need to deduce it from the given interval.
Since this interval is for the population mean, it is constructed arround the sample mean. That means, that the sample mean is at the center of the interval.
x[bar] = (Li+Ls)/2 ⇒ x[bar] = (13.62+15.89)/2 = 14.76%
The standard deviation needs a little more calculation,
amplitude a= Ls - Li ⇒ a= 15.89-13.62= 2.27
semiamplitude d= a/2 ⇒ d=2.27/2= 1.135lb ≅ 1.14
⇒
= 1.96
d= * (S/√n) ⇒ 1.14= 1.96 * (S/√131)
⇒ S= 6.657% ≅ 6.66%
Now the new CI has a confidence level of 99%
The Z value you need to use is = 2.58
x[bar] ± * (S/√n) ⇒ 14.76 ± 2.58 * (6.66/√131)
⇒ [13.25;16.26]%
With a confidence level of 99% you can expect that the interval [13.25;16.26]% contains the population mean of the backpack weight as percentage of body weight of sixth graders.
C.
"The American Academy of Orthopedic Surgeons recommends that backpack weight is at most 10% of body weight." As you can see in the calculated interval [13.25;16.26]%, the weight carried by the sixth graders is above the recommended backpack weight as percentage of body weight. This means that the sixth graders carry more weight than recommended.
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