In a random sample of 500 college students, 23% say that they read or watch the news every day. Develop a 90% confidence interval for the proportion of all students who read or watch the news on a daily basis. Interpret your results. If you wanted to develop a 95% confidence interval with a margin of error of .01, how many students would need to be surveyed?

Answers

Answer 1
Answer:

Answer:

The 90% confidence interval is  0.199 <  p < 0.261

The sample size to develop a 95% confidence interval is n = 2032  

Step-by-step explanation:

From the question we are told that

   The sample size is n =500

    The sample proportion is  \^ p = 0.23

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of  (\alpha )/(2) is  

   Z_{(\alpha )/(2) } =  1.645

Generally the margin of error is mathematically represented as  

     E =  Z_{(\alpha )/(2) } * \sqrt{(\^ p (1- \^ p))/(n) }

=>   E =  1.645 * \sqrt{(0.23 (1- 0.23))/(500) }

=>   E =  0.03096

Generally 90% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

=>    0.23  -0.03096  <  p < 0.23  +  0.03096

=>   0.199 <  p < 0.261

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  (\alpha )/(2) is  

   Z_{(\alpha )/(2) } =  1.96

The margin of error is given as E = 0.01

Generally the sample size is mathematically represented as  

    n = [\frac{Z_{(\alpha )/(2) }}{E} ]^2 * \^ p (1 - \^ p )

=>    n = [(1.96 )/(0.01) ]^2 *0.23 (1 - 0.23 )      

=>    n = 2032  


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Answers

Answer:

0.8809

Step-by-step explanation:

Given that:

The population proportion p = 4% = 4/100 = 0.04

Sample mean x = 16

The sample size n = 300

The sample proportion \hat  p =(x)/(n)

= 16/300

= 0.0533

P(\hat p \leq 0.0533) = P\bigg ( \frac{\hat p - p}{\sqrt{(P(1-P))/(n)}}\leq\frac{0.0533 - 0.04}{\sqrt{(0.04(1-0.04))/(300)}}\bigg )

P(\hat p \leq 0.0533) = P\bigg ( Z\leq\frac{0.0133}{\sqrt{(0.04(0.96))/(300)}}\bigg )

P(\hat p \leq 0.0533) = P\bigg ( Z\leq\frac{0.0133}{\sqrt{(0.0384)/(300)}}\bigg )

P(\hat p \leq 0.0533) = P\bigg ( Z\leq\frac{0.0133}{\sqrt{1.28 * 10^(-4)}}\bigg )

P(\hat p \leq 0.0533) = P\bigg ( Z\leq(0.0133)/(0.0113)}\bigg )

P(\hat p \leq 0.0533) = P\bigg ( Z\leq1.18}\bigg )

From the z tables;

= 0.8809

OR

Let X be the random variation that follows a normal distribution;

Then;

population mean \mu = n × p

population mean \mu = 300 × 0.04

population mean \mu = 12

The standard deviation \sigma = √(np(1-p))

The standard deviation  \sigma = √(300 * 0.04(1-0.04))

The standard deviation \sigma = √(11.52)

The standard deviation  \sigma = 3.39

The z -score can be computed as:

z = (x - \mu)/(\sigma)

z = (16 -12)/(3.39)

z = (4)/(3.39)

z = 1.18

The required probability is:

P(X ≤ 10) = Pr (z  ≤ 1.18)

= 0.8809

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Answers

Answer: pontos livres

Step-by-step explanation:

o cachorrinho wwss veyry rápido

Will give 100 points to anyone who can find the arc measure and show work! thanks!​

Answers

Answer:

  • 2π/3 radians

----------------------

Given:

  • Arc length (s) = 8π/3
  • Radius (r) = 4 km

To find the arc measure (θ) in radians, we can use the formula:

  • s = rθ

Now, plug in the given values:

  • 8π/3 = 4θ

To solve for θ, divide both sides by 4:

  • (8π/3) / 4 = θ

Simplify the expression:

  • 2π/3 = θ

So, the arc measure (θ) is 2π/3 radians