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3 years ago

12

I keep getting it wrong so help pls

1 answer:

bezimeni [28]3 years ago

4 0

The correct answer is 9000. That is because the digit 9 is in the thousands place.

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Two random samples are taken, with each group asked if they support a particular candidate. A summary of the sample sizes and pr

Minchanka [31]

Answer:

And we got $\alpha/2 =0.01$ so then the value for $\alpha=0.02$ and then the confidence level is given by: $Conf=1-0.02=0.98[/tex[ or 98%Step-by-step explanation:A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval". The margin of error is the range of values below and above the sample statistic in a confidence interval. Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean". [tex]p_1$ represent the real population proportion for 1

$\hat p_1 =0.768$ represent the estimated proportion for 1

$n_1=92$ is the sample size required for 1

$p_2$ represent the real population proportion for 2

$\hat p_2 =0.646$ represent the estimated proportion for 2

$n_2=95$ is the sample size required for 2

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

For this case we have the confidence interval given by: (-0.0313,0.2753). From this we can find the margin of erro on this way:

$ME= \frac{0.2753-(-0.0313)}{2}=0.1533$

And we know that the margin of erro is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

We have all the values except the value for $z_{\alpha/2}$

So we can find it like this:

$0.1533=z_{\alpha/2} \sqrt{\frac{0.768(1-0.768)}{92} +\frac{0.646 (1-0.646)}{95}}$

And solving for $z_{\alpha/2}$ we got:

$z_{\alpha/2}=2.326$

And we can find the value for $\alpha/2$ with the following excel code:

"=1-NORM.DIST(2.326,0,1,TRUE)"

And we got $\alpha/2 =0.01$ so then the value for $\alpha=0.02$ and then the confidence level is given by: $Conf=1-0.02=0.98$ or 98%

7 0

3 years ago

6 cm<br>What is the area of this shape?<br>5 cm<br>8 cm

erica [24]

Where is the shape or do you have a picture of it

7 0

3 years ago

For the following rectangular prism, find the length of the diagonal connecting A to B. Round to the nearest tenth.

meriva

Answer:

I think it is 192

Step-by-step explanation:

Volume of cube: 5 x 4 x 8 = 160

Area of diagonal: 160/2 = 80

Length of diagonal: 80 / 0.5 x 3 = 480

AB = 480 / 5 x 2 = 192

8 0

2 years ago

Solve for c.<br><br> 5c+16.5=13.5+10c

Kruka [31]

Answer:

c = 3/5

Step-by-step explanation:

5c+16.5=13.5+10c

Subtract 5c from each side

5c-5c+16.5=13.5+10c-5c

16.5 = 13.5 +5c

Subtract 13.5 from each side

16.5 -13.5 =13.5-13.5 +5c

3 = 5c

Divide by 5

3/5 = 5c/5

3/5 =c

4 0

4 years ago

Read 2 more answers

I need clause to this : 64-(13-7)

katen-ka-za [31]

64 - (13 - 7) = 64 - 6 = 58

7 0

4 years ago

Read 2 more answers