Cameron completed the following long division problem. Identify the error in his division.

What is greater 0.83 or 8.3

Grace [21]

8.3 is 10 times 0.83 so it is greater by a factor of 10

3 0

3 years ago

If g(x) = x3 + 4x2 + 5x + 2, which statement is true?

gtnhenbr [62]

Which statements, do you have any pics of the statements?

4 0

3 years ago

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Vanessa exercised by walking around a pond. One day, she walked 3 3/8 mi. The distance around the pond is 3/8 mi.

REY [17]

Let n be the number of times she walked around the pond

(3 3/8) / (3/8) = n

=> 9 = n

Vanessa walked around the pond 9 times

7 0

3 years ago

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My project has me plotting points on an uneven paper. How do I do this?

AURORKA [14]

You can make up your own lines on the paper or just get a new sheet of blank paper or graphing paper

4 0

3 years ago

Read 2 more answers

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

ZanzabumX [31]

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: $p\geq 0.41$

4 0

2 years ago