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icylaila
12.11.2020 •
Mathematics
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?
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Ответ:
The 90% confidence interval is![0.199 < p < 0.261](/tpl/images/0891/5199/cdd8b.png)
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![\^ p = 0.23](/tpl/images/0891/5199/0dc7c.png)
From the question we are told the confidence level is 90% , hence the level of significance is
=>![\alpha = 0.10](/tpl/images/0891/5199/b540d.png)
Generally from the normal distribution table the critical value of
is
Generally the margin of error is mathematically represented as
=>![E = 1.645 * \sqrt{\frac{0.23 (1- 0.23)}{500} }](/tpl/images/0891/5199/bd12a.png)
=>![E = 0.03096](/tpl/images/0891/5199/e256a.png)
Generally 90% confidence interval is mathematically represented as
=>![0.23 -0.03096 < p < 0.23 + 0.03096](/tpl/images/0891/5199/23438.png)
=>![0.199 < p < 0.261](/tpl/images/0891/5199/cdd8b.png)
From the question we are told the confidence level is 95% , hence the level of significance is
=>![\alpha = 0.05](/tpl/images/0891/5199/246a1.png)
Generally from the normal distribution table the critical value of
is
The margin of error is given as![E = 0.01](/tpl/images/0891/5199/e8d64.png)
Generally the sample size is mathematically represented as
=>
=>
Ответ:
and for 38min bike 15mph
and that will put him at 600.5 calories every hour