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Sergio [31]
3 years ago
14

428 students were surveyed about their preferences of sports. 146 students like football, 144 students like baseball, and 45 stu

dents like both sports. how many students like exactly one of the two sports?
Mathematics
1 answer:
ella [17]3 years ago
3 0

This is what we already know:


\left[\begin{array}{cccc} &YB&NB&total\\YF&45&[unknown]&146\\NF&[unknown]&[unknown]&[unknown]\\total&144&[unknown]&428\end{array}\right]


And using this information, we can fill the table out. Remember that everything must add up to the total in the column and row. The complete table looks like this:


\left[\begin{array}{cccc} &YB&NB&total\\YF&45&101&146\\NF&99&183&282\\total&144&284&428\end{array}\right]


Using the complete table, we see the fraction for students that only like baseball is \frac{99}{428} and students who only like football is \frac{101}{428} . All you need to do is add these fractions together to get your answer.


\frac{99}{428} +\frac{101}{428} =\frac{200}{428}


In short, 200 students only like one of the two sports.

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Answer:

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c) Standardized score for the reported percentage using a sample size of 400 = 2.02

Since, most of the variables in a normal distribution should fall within 2 standard deviations of the mean, a sample mean that corresponds to standard deviation of 2.02 from the population mean makes it seem very plausible that the people that participated in this sample weren't telling the truth. At least, the mathematics and myself, do not believe that they were telling the truth.

Step-by-step explanation:

The mean of this sample distribution is

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But the sample mean according to the population mean should have been

Sample mean = population mean = nP

= 0.56 × 1600 = 896.

To find the interval of values where the sample proportion should fall 68%, 95%, and almost all of the time, we obtain confidence interval for those confidence levels. Because, that's basically what the definition of confidence interval is; an interval where the true value can be obtained to a certain level.of confidence.

We will be doing the calculations in sample proportions,

We will find the confidence interval for confidence level of 68%, 95% and almost all of the time (99.7%).

Basically the empirical rule of 68-95-99.7 for standard deviations 1, 2 and 3 from the mean.

Confidence interval = (Sample mean) ± (Margin of error)

Sample Mean = population mean = 0.56

Margin of Error = (critical value) × (standard deviation of the distribution of sample means)

Standard deviation of the distribution of sample means = √[p(1-p)/n] = √[(0.56×0.44)/1600] = 0.0124

Critical value for 68% confidence interval

= 0.999 (from the z-tables)

Critical value for 95% confidence interval

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Critical values for the 99.7% confidence interval = 3.000 (also from the z-tables)

Confidence interval for 68% confidence level

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Confidence interval for 95% confidence level

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c) Now suppose that the sample had been of only 400 people. Compute a standardized score to correspond to the reported percentage of 61%. Comment on whether or not you believe that people in the sample could all have been telling the truth, based on your result.

The new standard deviation of the distribution of sample means for a sample size of 400

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The standardized score for any is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (0.61 - 0.56)/0.0248 = 2.02

Standardized score for the reported percentage using a sample size of 400 = 2.02

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Hope this Helps!!!

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