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Mkey [24]
2 years ago
6

Students in Mrs. Curry's class recorded the amount of snowfall in January and February.

Mathematics
1 answer:
Mila [183]2 years ago
7 0

Answer:

A and C

Step-by-step explanation:

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Eeeeeeeee eeeeeee e e e e e fr jsdf jdfrjae reeeee?
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Answer:

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Step-by-step explanation:

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4 0
3 years ago
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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
1. What would be the best graph or display to represent your Home Library Statistics data? Why?
algol13

Answer:

A bar graph

Step-by-step explanation:

A bar chart will give you the visual understanding of your data.It will clearly show the book with most number of pages and compare the books according to the number of pages.You can formart using different colors for each bar.

4 0
3 years ago
4 cats each ate 1/4 cup of canned food and 1/4 cup of dry food. how much food did they eat altogether
VashaNatasha [74]

Answer: 1 cup canned and 1 cup dry, 2cups put together

Step-by-step explanation:

8 0
3 years ago
PLZZ HELPP YOU WILL BE THE BRAINLEIST!!!!!!!
sasho [114]

i need a graph so i can see then i will solve it

4 0
3 years ago
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