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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

3 years ago

Rewrite the expressions below by completing the square.

Mrrafil [7]

Answer:

(x + 2)^2 - 1

Step-by-step explanation:

x^2 + 4x + 1

= (x^2 + 4x + 2 - 2 + 1)

= (x + 2)^2 - 1

3 0

2 years ago

The equation of line a is y=12x−1, and it passes through point (6,2).

Triss [41]

Answer:

A). slope = -1/12 y-intercept = 3/2

Step-by-step explanation:

y = -1/12x + b

2 = -1/12(-6) + b

2 = 1/2 + b

3/2 = b

y = -1/12x + 3/2

3 0

2 years ago

What is and equation line that passes through the point (8,-5) and is parallel to the line 5x+4y=24

GenaCL600 [577]

Answer:

5x + 4x = 20

Step-by-step explanation:

You have your x and y values and if you substitute them in you find yourself with 40 - 20 = 24 this doesn't make sense. So you get another answer which can be set to 20 and 40 - 20 = 20 therefore you you can make sure that your line intersects with your x and y values.

4 0

3 years ago

Write the linear equation in slope intercept form 2x+By=16

allochka39001 [22]

Answer:

Part 1) $y=-\frac{2x}{B}x+\frac{16}{B}$