Question

1.) A Retirement Confidence Survey of 1153 workers and retirees in the USA 25 years of age and older condcuted by Employee Benefit Research Institute in January 2010 found that 496 had less than $10,000 in Savings. Construct a 95% confidence interval for the proportion of USA workers and retirees 25 years old or older who have less than $10,000 in savings. USE PANIC 2.) A random sample of 1003 adult Americans was asked, “Do you pretty much televisions are a necessity or luxury you could do without?" Pf the 1003 adults surveyed, 521 indciated that televisions are a luxury they could do without. Construct a 95% confidence interval for the proportion of adult Americans who believe that television are a luxury they could do without USE PANIC

PANIC = prameter , assumptions , name, interval, conclusion

Answer 1

Answer:

1.

Given,

sample proportion p^ = x/n

substitute values

= 496/1153

= 0.4302

Here at 95% CI, z value is 1.96

95% CI = p^ +/- z*sqrt(p^q^/n)

substitute values

= 0.4302 +/- 1.96*sqrt(0.4302(1-0.4302)/1153)

= 0.4302 +/- 0.0286

= (0.4016 , 0.4588)

95% of the time in the population 40.16% to 45.88% of the workers and retires in the US 25 years of age and older have less than $10000 in savings.

2.

Given,

sample proportion p^ = x/n

substitute values

= 521/1003

= 0.5194

Here at 95% CI, z value is 1.96

95% CI = p^ +/- z*sqrt(p^q^/n)

substitute values

= 0.5194 +/- 1.96*sqrt(0.5194(1-0.5194)/1003)

= 0.5194 +/- 0.0309

= (0.4885 , 0.5503)

95% interval lies b/w 48.85% to 55.03%