A simple random sample of 150 people who completed the ESPN 2019 Tournament Bracket was selected, and 81 out of the 150 people who completed the ESPN 2019 Tournament Bracket had their choice to become the national champion make the Final Four. If appropriate, use this information to calculate and interpret a 90% confidence interval for the proportion of all people who completed the ESPN 2019 Tournament Bracket whose choice to become the national champion made the Final Four.

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Answer:

The answer is "0.54 and (0.4733, 0.6067)".

Step-by-step explanation:

Please find the complete question in the attachment file.

In question 1:

[tex]\to x=81\\\\ \to n=150\\\\ \to \hat{p}=\frac{x}{n}=\frac{81}{150}=0.54\\\\[/tex]

Hence the choice A is correct.

In point b:

Calculating Critical Value:

[tex]\to (Z_{\alpha/2})=1.64 \ \ \ \ \ \ \ \ \ \ (use \ \ z\ \ table)\\\\[/tex]

Calculating the Confidence intervals:

[tex]=\hat{p} \pm Z_{\alpha/2}\times \sqrt{\frac{\hat{p} \times (1-\hat{p}))}{n}}\\\\ =0.54 \pm 1.64\times \sqrt{\frac{0.0.54 \times (1-0.54)}{150}}\\\\ =0.54 \pm 0.0667\\\\ =(0.4733, 0.6067)\\\\[/tex]

Hence the choice a is correct.

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The correct confidence interval is (0.4733, 0.6067).

What is the z critical value?

The z critical test is conducted when the population standard deviation on a normal distribution is known and the sample size is greater than or equal to 30.

x = 81

n = 150

p = x/n =81/ 150

p  = 0.54

So, the correct choice is A.

In point b

Critical Value:

[tex]Z_{\alpha /2} = 1.64[/tex]

the Confidence intervals:

= p±[tex]Z_{\alpha /2} \times \sqrt{\frac{p \times (1-p)}{n} }[/tex]

= 0.54 ± [tex]\\1.64 \times \sqrt{\frac{0.54 \times (1-0.54)}{150} }[/tex]

= 0.54 ± 0.0667

= (0.4733, 0.6067)

Hence, the choice A is correct.

Therefore, the correct confidence interval is (0.4733, 0.6067).

Learn more about critical value:

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