Is 1\8 equivalent to 4\32

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

Is 4y^2*3y^5=12y^10 true or false.

mixer [17]

Answer:False

Step-by-step explanation:#UwU God

5 0

3 years ago

Find the range. 4.7 6.3 5.4 3.2 4.9 3.1 –3.1 9.5 –9.5

WINSTONCH [101]

Answer:

The range is 19

Step-by-step explanation:

First, you have to order the numbers from least to greatest:

-9.5, -3.1, 3.1, 3.2, 4.7, 4.9, 5.4, 6.3, 9.5

Then, you have to find the difference between the largest number and and smallest number. You do this by subtracting them:

9.5 - (-9.5)= 19

So, the range is 19

8 0

3 years ago

How do you do this?????????????

uranmaximum [27]

Hey there!

To find the value of an angle in a triangle, you need to add the 3 angles. You have angles with 55 degrees, 65 degrees, and x degrees. Put this into an equation. The equation would be 55+65+x=180. Simplify to get 120+x=180. You need to isolate x, so subtract 120 from both sides to get x=60. So the value of x is 60 degrees.

I hope this helps!

7 0

4 years ago

Read 2 more answers

What is the slope of the line x = 4?

rewona [7]

The slope is undefined because you need a y intercept as well.

8 0

3 years ago