Question

In: Statistics and Probability

In an International Mass Retail Association survey of 1006 U.S. households, 324 of the households said...

In an International Mass Retail Association survey of 1006 U.S. households, 324 of the households said that Saturday is their favorite shopping day (USA Today, October 6, 1993). Construct a 95% confidence interval for the proportion of all U.S. households for whom Saturday is their favorite shopping day. Round your answer to three decimal places.

Expert Solution

Solution :

Given that,

n = 1006

x = 324

Point estimate = sample proportion = = x / n = 324/1006=0.322

1 - = 0.678

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

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

= 1.96 ( $\sqrt$ ((0.322*0.678) /1006 )

E = 0.029

A 95% confidence interval is ,

- E < p < + E

0.322-0.029 < p < 0.322+0.029

0.293< p < 0.351

orchestra answered 2 years ago

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