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Anvisha [2.4K]
4 years ago
8

PLEASE HELP!!! PLEASE HURRYYY!!! I'm on timed quiz!!

Mathematics
1 answer:
miss Akunina [59]4 years ago
8 0

the slope is -2 and the y-intercept is 12

Step-by-step explanation:

1. choose two ordered pairs from the table (x1,y1) and (x2,y2).

2. write down the slope formula

3. replace the xs and ys by their values

4. subtract

5. divide

1. to solve for the y intercept, choose any one of the ordered pairs

2. replace the letters by their values

3. multiply

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A sample of 200 observations from the first population indicated that x1 is 170. A sample of 150 observations from the second po
igor_vitrenko [27]

Answer:

a) For this case the value of the significanceis \alpha=0.05 and \alpha/2 =0.025, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

z_{\alpha/2} =1.96

If the calculated statistic |z_{calc}| >1.96 we can reject the null hypothesis at 5% of significance

b) Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8  

c)z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708    

d) Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

Step-by-step explanation:

Data given and notation    

X_{1}=170 represent the number of people with the characteristic 1

X_{2}=110 represent the number of people with the characteristic 2  

n_{1}=200 sample 1 selected  

n_{2}=150 sample 2 selected  

p_{1}=\frac{170}{200}=0.85 represent the proportion estimated for the sample 1  

p_{2}=\frac{110}{150}=0.733 represent the proportion estimated for the sample 2  

\hat p represent the pooled estimate of p

z would represent the statistic (variable of interest)    

p_v represent the value for the test (variable of interest)  

\alpha=0.05 significance level given  

Concepts and formulas to use    

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:    

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

We need to apply a z test to compare proportions, and the statistic is given by:    

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

a.State the decision rule.

For this case the value of the significanceis \alpha=0.05 and \alpha/2 =0.025, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

z_{\alpha/2} =1.96

If the calculated statistic |z_{calc}| >1.96 we can reject the null hypothesis at 5% of significance

b. Compute the pooled proportion.

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8  

c. Compute the value of the test statistic.                                                                                              

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.    

Replacing in formula (1) the values obtained we got this:    

z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708    

d. What is your decision regarding the null hypothesis?

Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

5 0
3 years ago
A pair of grant high school hockey fans counted the number of games won by the school each year
Mrrafil [7]

Answer:

8

Step-by-step explanation:

2015: 9 games won

2014: 1 game won

rate of change = 2015 games won - 2014 games won

rate of change = 9 - 1

rate of change = 8

5 0
3 years ago
Horn lengths of Texas longhorn cattle are normally distributed. The mean horn spread is 60 inches with a standard deviation of 4
german

Answer:

range is  between 55.5 to 64.5

Step-by-step explanation:

Horn lengths of Texas longhorn cattle are normally distributed. The mean horn spread is 60 inches with a standard deviation of 4.5 inches

68% is 1 standard deviation from mean

To get the range of 1 standard deviation we add and subtract standard deviation from mean

mean = 60

standard deviation = 4.5

60 - 4.5= 55.5

60+4.5 = 64.5

1 standard deviation is between 55.5 to 64.5

That is 68% range is  between 55.5 to 64.5

8 0
3 years ago
Why did delegates argue over representation in congress?
KatRina [158]
They argued because they didnt agree over the congresses conclusion they made with the delegates.
5 0
3 years ago
HELP ME QUICK!!! this has to be turned in in 46 min
IRISSAK [1]

Answer:

the answer is most likely A

Step-by-step explanation:

4 0
3 years ago
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