WORTH 30 POINTS! Replace ... with <, ≤, =, >, or ≥ so that the inequality will be true for any value of x.

Mathematics

2 answers:

Len [333]1 year ago

8 0

Hope this helps a bit

hichkok12 [17]1 year ago

6 0

The other persons is right:D

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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

A square has a side of 6.25 feet. What is the area in square feet? Round to the correct significant digits.

denis23 [38]

Area of a square = side^2
area = 6.25^2 = 39.0625 square feet

= 39.06 (rounded)

4 0

2 years ago

I'LL GIVE BRAINLIEST TO WHO EVER ANSWERS!!!!

Nookie1986 [14]

Answer:

Answer is Step 2

6 0

2 years ago

Mr Torres has seven dollars in $1 bill and fifteen dollars in $5 bills. If he gives his daughter three bills that are not all th

adoni [48]

Two can be the same and for the most that would be two $5 bills and the third would be a $1 bill so the most he could give her is $11

4 0

2 years ago

The price of oil, in dollars per barrel, declined last week by 3.5%. If it started the week at 102.00 per barrel, at what per ba

scoray [572]

Answer:

98.57

Step-by-step explanation:

3.5% over 100% X 102=3.43

102-3.43= 98.57

6 0

3 years ago