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?

The 90% confidence interval is  

The sample size to develop a 95% confidence interval is  

Step-by-step explanation:

From the question we are told that

   The sample size is n =500

    The sample proportion is  

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

=>  

Generally from the normal distribution table the critical value  of   is  

Generally the margin of error is mathematically represented as  

=>  

=>  

Generally 90% confidence interval is mathematically represented as  

=>    

=>  

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

=>  

Generally from the normal distribution table the critical value  of   is  

The margin of error is given as

Generally the sample size is mathematically represented as  

=>          

=>      

b 1: 45 am

step-by-step explanation:

its that because if it was in the pm it would be 1345

d) 30 units^2

area of rectangle = l × b

we know,

length = 10 unit

breadth = 3 unit

applying the formula,

= l × b

= 10 × 3

= 30 unit^2

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