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Mkey [24]
3 years ago
10

A survey asked a group of students to choose their favorite type of sport from the choices of soccer, softball, basketball, and

others. The results of the survey are shown in the graph. Based on the graph, how many students in a class of 84 students would be expected to choose a sport other than soccer, softball, or basketball as their favorite type of sport?
Mathematics
1 answer:
jolli1 [7]3 years ago
3 0
The answer would be 12 students. 8+4+6+3=21. 
     84 divided by 21 would = 4. 
  3 students times 4 = 12 students
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Answer:

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

And the confidence interval is given by:

(0.123, 0.177)

And for this case the interval contains the value 0.16, so then we can conclude at 5% of significance that the true proportion is not different from 0.16

Step-by-step explanation:

Previous concepts

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".  

The population proportion have the following distribution

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

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

And the confidence interval is given by:

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And for this case the interval contains the value 0.16, so then we can conclude at 5% of significance that the true proportion is not different from 0.16

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