Correct answer only!!!

Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8

Delvig [45]

We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51$

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98$

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34$
Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86$

Thus the 98% confidence interval will be (105.34, 141.86)

Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.

3 0

4 years ago

124 - 8g = -172 <br> Find (g).

SCORPION-xisa [38]

The answer is g=37. This is correct

6 0

4 years ago

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A line l2 is perpendicular to the line x + 2y – 5 = 0 without computing the gradient of this

Stells [14]

Answer:

y=2x-2

Step-by-step explanation:

first, let's put the line into y=mx+b form (slope-intercept form), where m is the slope (or gradient) and b is the y intercept

add 5 to both sides

x+2y=5

subtract x from both sides

2y=-x+5

divide by 2

y=-1/2x+5/2

perpendicular lines have slopes (gradients) that are negative (one is positive, another one is negative) and reciprocal (they are essentially the same number, just "flipped")

to find the slope of I2:

since -1/2 is negative, that means the slope of I2 will be 2 (2/1 is the reciprocal of 1/2 (positive version of -1/2))

so here's our equation so far:

y=2x+b

now we need to find b

because line will pass through (3,4), we can use it to solve for b

substitute 4 as y and 3 as x

4=2(3)+b

multiply

4=6+b

-2=b

therefore the equation is y=2x-2

hope this helps!!

3 0

3 years ago

I Need a lot of help

Darina [25.2K]

Answer:

i think B if i am wrong i will give you points

Step-by-step explanation:

:)

3 0

3 years ago

What is the length of each segment?<br> Segment AB: from A(-2,1) to B(-2, 5)

White raven [17]

Answer:

$length = \sqrt{ {( - 2 - ( - 2))}^{2} + {(5 - 1)}^{2} } \\ = \sqrt{16} \\ = 4 \: units$

5 0

3 years ago

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