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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

Respuesta :

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. 
InesWalston

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,

[tex]=p\pm z\cdot \text{Standard Error}[/tex]

where,

z = Z score of the confidence interval,

Standard Error = [tex]\sqrt\dfrac{p(1-p)}{n}[/tex]

Putting the values in the standard error formula,

Standard Error,

[tex]=\sqrt\dfrac{0.8(1-0.8)}{49}[/tex]

[tex]=\sqrt\dfrac{0.8(0.2)}{49}[/tex]

[tex]=0.0571[/tex]

For a 95% confidence interval, z=1.96 and

[tex]=0.8\pm 1.96\cdot 0.0571[/tex]

[tex]=0.8\pm 0.112[/tex]

[tex]=0.688,0.912[/tex]

[tex]=68\%,92\%[/tex]

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