Question

Suppose that a simple random sample of 125 people finds that 43 eat chocolate every day. Use this information to answer the following. Step 4 of 5: Find the upper bound for a 98% confidence interval of the true proportion of people who eat chocolate daily, accurate to 3 decimal places.

Answer 1

Solution :

Given that,

n = 125

x = 43

= x / n = 43 /125 = 344

1 - = 1 - 0.344 = 0.656

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.326 * (((0.344 * 0.656 ) / 125 )

= 0.099

A 98 % confidence interval for population proportion p is ,

+ E

0.344 + 0.099

The upper bound = 0.443