Question

In: Statistics and Probability

Kate’s survey showed that 54 of 120 employees eat fast food at least once a week...

Kate’s survey showed that 54 of 120 employees eat fast food at least once a week for lunch. What is the proportion of employees who eat fast food at least once a week, using a 98% confidence interval?

Group of answer choices

24% to 44%

30% to 50%

34% to 56%

28% to 45%

Expert Solution

Solution :

Given that,

Point estimate = sample proportion = = x / n = 54/120=0.45

1 - = 1-0.45 =0.55

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table )

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

= 2.326 ( $\sqrt$ ((0.45*0.55) /120 )

E = 0.11

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.45-0.11 < p < 0.45+0.11

0.34< p < 0.56

34% to 56%

orchestra answered 2 years ago

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