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IrinaVladis [17]
1 year ago
10

Concerned about graffiti, mayors of nine suburban communities instituted a citizen community watch program. community monthly in

cidents after monthly incidents before burr oak 2 7 elgin corners 3 2 elm grove 6 2 greenburg 4 3 huntley 5 11 north lyman 3 9 south lyman 3 5 pin oak 5 7 victorville 2 3 click here for the excel data file (a) choose the appropriate hypotheses to see whether the number of graffiti incidents declined. assume μd is the mean difference in graffiti incidents before and after.

Mathematics
2 answers:
Likurg_2 [28]1 year ago
7 0
This is a problem on hypothesis testing on the difference of two dependent means. To solve for this one, the difference for each pair should be solved. 

After it, the mean and standard deviation of the differences should also be calculated.  Then, we state the null and alternative hypotheses. 

The null hypothesis is  H_{0}: \mu_d=0
The alternative hypothesis is  H_a: \mu_d\ \textgreater \ 0

These are the required hypotheses <span>to see whether the number of graffiti incidents declined. </span>
Aleonysh [2.5K]1 year ago
7 0
In order to be clear, the data are reported in the picture attached.

The null hypothesis H₀ represents what is believed true if the data don't prove the opposite.
The alternative hypothesis H₁ is the conclusion we want to deduce from the test.

If we call d the difference in graffiti before (B) and after (A) (pay attention to the order) d = B - A.
We want to prove that the incidents are declined, therefore B should be greater than A and d would be positive.

Therefore:
H₀ : μd ≤ 0
H₁ : <span>μd > 0 </span>

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In the figure CD is the perpendicular bisector of AB . If the length of AC is 2x and the length of BC is 3x - 5 . The value of x
Dennis_Churaev [7]

Answer

Find out the value of x .

To proof

SAS congurence property

In this property two sides and one angle of the two triangles are equal.

in the Δ ADC and ΔBDC

(1) CD = CD (common side of both the triangle)

(2) ∠CDA = ∠ CDB = 90 °

( ∠CDA +∠ CDB = 180 ° (Linear pair)

as given in the diagram

∠CDA  = 90°

∠ CDB = 180 ° - 90°

∠ CDB = 90°)

(3) AD = DB (as shown in the diagram)

Δ ADC ≅ ΔBDC

by using the SAS congurence property .

AC = BC

(Corresponding sides of the congurent triangle)

As given

the length of AC is 2x and the length of BC is 3x - 5 .

2x = 3x - 5

3x -2x =5

x = 5

The value of x is 5 .

Hence proved


7 0
1 year ago
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It has been suggested that night shift-workers show more variability in their output levels than day workers. Below, you are giv
bonufazy [111]

Answer:

Null hypotheses = H₀ = σ₁² ≤ σ₂²

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic = 1.9

p-value = 0.206

Since the p-value is greater than α therefore, we cannot reject the null hypothesis.

So we can conclude that the night shift workers don't show more variability in their output levels than day workers.

Step-by-step explanation:

Let σ₁² denotes the variance of night shift-workers

Let σ₂² denotes the variance of day shift-workers

State the null and alternative hypotheses:

The null hypothesis assumes that the variance of night shift-workers is equal to or less than day-shift workers.

Null hypotheses = H₀ = σ₁² ≤ σ₂²

The alternate hypothesis assumes that the variance of night shift-workers is more than day-shift workers.

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic:

The test statistic or also called F-value is calculated using

Test statistic = Larger sample variance/Smaller sample variance

The larger sample variance is σ₁² = 38

The smaller sample variance is σ₂² = 20

Test statistic = σ₁²/σ₂²

Test statistic = 38/20

Test statistic = 1.9

p-value:

The degree of freedom corresponding to night shift workers is given by

df₁ = n - 1

df₁ = 9 - 1

df₁ = 8

The degree of freedom corresponding to day shift workers is given by

df₂ = n - 1

df₂ = 8 - 1

df₂ = 7

We can find out the p-value using F-table or by using Excel.

Using Excel to find out the p-value,

p-value = FDIST(F-value, df₁, df₂)

p-value = FDIST(1.9, 8, 7)

p-value = 0.206

Conclusion:

p-value > α    

0.206 > 0.05   ( α = 1 - 0.95 = 0.05)

Since the p-value is greater than α therefore, we cannot reject the null hypothesis corresponding to a confidence level of 95%

So we can conclude that the night shift workers don't show more variability in their output levels than day workers.

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Answer:

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Step-by-step explanation:

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25 minutes

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