Write an equation for the line in slope-intercept form

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

Y + 8 = 2 i need help!!!!!

sergey [27]

Answer:

y = -6

Step-by-step explanation:

-6 + 8 = 2

3 0

3 years ago

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G ^-1(3)= A.-1 B.0 C.3 D. 5 Please help me understand this!!

Natasha2012 [34]

The answer the 3,

First you have to apply the exponent rule which is 1 - 3• 1/g

Then you multiply the fractions - 1•3/g

Then multiply the numbers 1 and 3, which is 3, so there you have it 3/g, or 3, they’re the same thing.

8 0

3 years ago

To see how costly it would be for other people to compete a similar cycling tour, you decide to work out two averages and compar

Damm [24]

Group A

Money spent by Person 1 = £500

Money spent by Person 2 = £600

Money spent by Person 3 = £900

Money spent by Person 4 = £450

Sum of money spent by group A = £2,450

Average money spent by group A = $\frac{2,450}{4}$

⇒ Average money spent by group A = £612.5

Group B

Money spent by Person 1 = £700

Money spent by Person 2 = £500

Money spent by Person 3 = £680

Money spent by Person 4 = £500

Sum of money spent by group B = £2,380

Average money spent by group B = $\frac{2,380}{4}$

⇒ Average money spent by group B = £595

Hence, Group B had lower average spend.

4 0

3 years ago

From point A(3,-2), walk 4 spaces in the negative x direction, and 3 spaces in the positive y direction. Use a paper cutout to l

Zolol [24]

Answer: 3,-2Step-by-step explanation: ... Use a paper cutout to label the point you reach as. B. What are the coordinates of point B?

Step-by-step explanation:

6 0

2 years ago

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