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kicyunya [14]
3 years ago
13

ducational Television In a random sample of people, said that they watched educational television. Find the confidence interval

of the true proportion of people who watched educational television. Round the answers to at least three decimal places.
Mathematics
1 answer:
SIZIF [17.4K]3 years ago
3 0

Complete Question

Educational Television In a random sample of 200 people, 159 said that they watched educational television. Find the 90% confidence interval of the true proportion of people who watched educational television. Round the answers to at least three decimal places.

Answer:

The 90% confidence interval is   0.748  <  p <  0.842      

Step-by-step explanation:

From the question we are told that

     The sample size is  n = 200

     The number of people that watched the educational television is k =  159

Generally the sample proportion is mathematically represented as

               \^ p =  \frac{159}{200}

=>               \^ p =  0.795

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  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the margin of error is mathematically represented as  

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

=>     E =  1.645  * \sqrt{\frac{0.795 (1- 0.795)}{200} }

=>     E = 0.04696

Generally 95% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

=>    0.795  -0.04696  <  p <  0.795 + 0.04696

=>    0.748  <  p <  0.842      

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