Question

A consumer surveys 49 televisions. 80% cost between $250.00 and $350.00. The administrator decides to calculate a 95% confidence interval for this proportion.
The confidence interval is from 68% to 92% (to the nearest percent).

True
False

Answer 1

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

$=0.8\pm 1.96\cdot 0.0571$

$=0.8\pm 0.112$

$=0.688,0.912$

$=68\%,92\%$

Answer 2

Thank you for posting your question here at brainly. I think the above statement is true as the confidence interval is from 68% to 92%. I hope the answer will help you.