A random sample of 1003 adults Americans was asked Do you pr

A random sample of 1003 adults Americans was asked \"Do you pretty much think televisions are a necessity or luxury you could do without ?\" Of 1003 adults sun eyed, 521 indicated that televisions are a luxury\' they could do without. Construct a 95% confidence interval for the population proportion of adult American who believe that television arc a luau they could do without. Determine the null and alternative hypothesis According to the National Association of home Builders the mean 222. A real estate broker Find the critical t-value that corresponds to 99% confidence and sample size n 15.

Solution

Q3.
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=521
Sample Size(n)=1003
Sample proportion = x/n =0.519
Confidence Interval = [ 0.519 ±Z a/2 ( Sqrt ( 0.519*0.481) /1003)]
= [ 0.519 - 1.96* Sqrt(0) , 0.519 + 1.96* Sqrt(0) ]
= [ 0.488,0.55]