Question

In: Statistics and Probability

In a random survey of 2221 adults aged 18 and older 689 of them said that...

In a random survey of 2221 adults aged 18 and older 689 of them said that football is their favorite sport to watch on television. Construct a 94% Confidence Interval for the true proportion of adults aged 18 and older for whom football is their favorite sport to watch on television.

A (.2909, .3295)

B (.2918, .3287)

C (.3185. .3401)

D (.3102, .3214)

Expert Solution

Solution :

n = 2221

x = 689

= x / n = 689 / 2221 = 0.3102

1 - = 1 - 0.3102 = 0.6898

At 94% confidence level the z is ,

= 1 - 94% = 1 - 0.94 = 0.06

/ 2 = 0.06 / 2 = 0.0

Z_/2 = Z_0.03 = 1.881

Margin of error = E = Z_{/ 2} * $\sqrt$ (( * (1 - )) / n)

= 1.881 * ( $\sqrt$ ((0.3102 * 0.6898) / 2221 )

= 0.0185

A 94% confidence interval for population proportion p is ,

- E < P < + E

0.3102 - 0.0185 < p < 0.3102 + 0.0185

0.2918 < p < 0.3287

( 0.2918,0.3287 )

Option B ) is correct.

orchestra answered 2 years ago

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