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Alisiya [41]
3 years ago
9

A car company developed a certain car model to appeal to young consumers. The car company claims the average age of drivers of t

his certain car model is 28.00 years old. Suppose a random sample of 17 drivers was​ drawn, and the average age of the drivers was found to be 29.00 years. Assume the standard deviation for the age of the car drivers to be 2.2 years.
Required:
a. Construct a 95% confidence interval to estimate the average age of the car driver.
b. Does this result lend support to the car's company's claim?
c. What assumption needs to be made to construct this's interval?
Mathematics
1 answer:
Klio2033 [76]3 years ago
6 0

Answer:

(27.954, 30.046)

Yes

Step-by-step explanation:

Given that :

Mean (m) = 29

Standard deviation (σ) = 2.2

Sample size (n) = 17

Confidence interval (α) = 95%

Average age as claimed by the company = 28

Confidence interval formula :

Mean ± (Zcritical * σ/sqrt(n))

Zcritical at 95% = 1.96

(Zcritical * σ/sqrt(n)) = (1.96 * (2.2/sqrt(17)) = (1.96 * 0.5335783) = 1.0458136

Lower boundary = (29 - 1.0458136) = 27.954186

Upper boundary = (29 + 1.0458136) = 30.045813

(27.954, 30.046)

b. Does this result lend support to the car's company's claim?

Yes, it does, 28 falls in between the the calculated confidence interval.

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x = (2, 7, 6, 10, 9, 13, 11, 18)

y = (11, 21, 12, 25, 7, 12, 4, 7)

a. The formula used for Sample Covariance is:

S_{XY}=\frac{\sum_{i=1}^{n}(X_{i}-\bar{X})(Y_{i}-\bar{Y})}{n-1}

where, \bar{X} is mean of X

and \bar{Y}  is mean of Y

Firstly calculating \bar{X} and \bar{Y}. We get,

\bar{X} = 9.5

\bar{Y} = 12.375

Now, Putting all values in above formula. We get,

Sample Covariance S_{XY} = -8.64

b. Standard Deviation is the square root of sum of square of the distance of observation from the mean.  

Standard deviation(\sigma) = \sqrt{\frac{1}{n}\sum_{i=1}^{n}{(x_{i}-\bar{x})^{2}} }

where, \bar{x} is mean of the distribution.

Using formula we get,

Standard deviation (\sigma_{X}) = 4.81

c. Using above formula we get,

Standard deviation (\sigma_{Y}) = 7.21

d. Correlation Coefficient is calculate by using formula:

r_{xy}=\frac{S_{XY}}{S_{X}S_{Y}}

where, S_{XY} = Covariance of X and Y

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S_{Y} = Standard Deviation of Y

Putting all values in above formula, we get,

Correlation Coefficient (r_{xy}) = -0.25

<em>Correlation Coefficient tell us that X and Y has negative week relationship whereas Sample Covariance tell us that X and Y has negative relationship. It does not tell us strength of relationship.</em>

5 0
3 years ago
Please help! Need answer ASAP!
Umnica [9.8K]

Answer:

Step-by-step explanation:

A) identical triangles PSR and PSQ

B) in the given triangle PQR angle P is 90 degrees and PS is perpendicular on QR.  so the angle PSR and angle PSQ is 90 degrees also angles RPS , SPQ are 45 degrees . hence when two angles of a triangle are 90 and 45 degrees.  the third is also 45 degrees. when the three angles of two triangles are equal they are said to be similar

C) the triangle PSR is a isosceles triangle therfore the two sides of the two oposite equal angles must be equal. if RS=4  then SP=4 as well the sides opposite to 45 degrees angles. the third side can be found by using the formula

c²=a²+b²

c²=4²+4²=16+16=32

c=√32=√16*2=4√2

4 0
3 years ago
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