Answer:
True.
Step-by-step explanation:
A consumer surveys 49 televisions. 80% cost between $250.00 and $350.00
So here,
p = proportion = 80% = 0.80,
n = sample size = 49,
We know that, confidence interval is,
\(=p\pm z\cdot \text{Standard Error}\)
where,
z = Z score of the confidence interval,
Standard Error = \(\sqrt\dfrac{p(1-p)}{n}\)
Putting the values in the standard error formula,
Standard Error,
\(=\sqrt\dfrac{0.8(1-0.8)}{49}\)
\(=\sqrt\dfrac{0.8(0.2)}{49}\)
\(=0.0571\)
For a 95% confidence interval, z=1.96 and
\(=0.8\pm 1.96\cdot 0.0571\)
\(=0.8\pm 0.112\)
\(=0.688,0.912\)
\(=68\%,92\%\)
