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almond37 [142]
3 years ago
7

2r - 3.1 =1.7 ; 4t + 3.5 =12.5 ; 8m - 5.5 =10.1 i really need help plzzzzzzz

Mathematics
1 answer:
IRISSAK [1]3 years ago
6 0
2r - 3,1 = 1,7

<span>Keep the numbers with incognito on the left and the numbers without incognito on the right. Exchanging the signs when changing sides if applicable.
</span>2r = 1,7 + 3,1
2r = 4,8
r =  \frac{4,8}{2}
\boxed{r = 2,4}


--------------------------------
4t + 3,5 = 12,5
<span>Keep the numbers with incognito on the left and the numbers without incognito on the right. Exchanging the signs when changing sides if applicable.
</span>4t = 12,5-3,5
4t = 9,0
t =  \frac{9,0}{4}
\boxed{t = 2,25}
-------------------------------------

8m - 5.5 = 10.1
<span>Keep the numbers with incognito on the left and the numbers without incognito on the right. Exchanging the signs when changing sides if applicable.
</span>8m = 10,1 + 5,5
8m = 15,6
m =  \frac{15,6}{8}
\boxed{m = 1,95}



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Based on the Nielsen ratings, the local CBS affiliate claims its 11:00 PM newscast reaches 41 % of the viewing audience in the a
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Answer:

1) Null hypothesis:p\geq 0.41  

Alternative hypothesis:p < 0.41

2) \hat p=0.36 estimated proportion of people indicated that they watch the late evening news on this local CBS station

3) z_{crit}=-2.33

And we can use the following excel code to find it: "=NORM.INV(0.01,0,1)"

4) z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{1000}}}=-1.017  

5) z_{crit}=-1.28

And we can use the following excel code to find it: "=NORM.INV(0.1,0,1)"

6) We see that |t_{calculated}| so then we have enough evidence to FAIL to reject the null hypothesis at 1% of significance.

7) Null hypothesis:p\geq 0.41  

Step-by-step explanation:

Data given and notation  

n=100 represent the random sample taken

X represent the people indicated that they watch the late evening news on this local CBS station

\hat p=0.36 estimated proportion of people indicated that they watch the late evening news on this local CBS station

p_o=0.41 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v{/tex} represent the p value (variable of interest)  Part 1We need to conduct a hypothesis in order to test the claim that 11:00 PM newscast reaches 41 % of the viewing audience in the area:  Null hypothesis:[tex]p\geq 0.41  

Alternative hypothesis:p < 0.41

Part 2  

\hat p=0.36 estimated proportion of people indicated that they watch the late evening news on this local CBS station

Part 3

Since we have a left tailed test we need to see in the normal standard distribution a value that accumulates 0.01 of the area on the left and on this case this value is :

z_{crit}=-2.33

And we can use the following excel code to find it: "=NORM.INV(0.01,0,1)"

Part 4

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{1000}}}=-1.017  

Part 5

Since we have a left tailed test we need to see in the normal standard distribution a value that accumulates 0.1 of the area on the left and on this case this value is :

z_{crit}=-1.28

And we can use the following excel code to find it: "=NORM.INV(0.1,0,1)"

Part 6

We see that |t_{calculated}| so then we have enough evidence to FAIL to reject the null hypothesis at 1% of significance.

Part 7

Null hypothesis:p\geq 0.41  

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Answer:

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150+55w

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