Which expression should you simplify to find the 90% confidence interval,
given a sample of 64 people with a sample proportion of 0.55?

Step-by-step explanation:

CI = p ± z √(pq/n)

For 90% confidence, z = 1.645.

CI = 0.55 ± 1.645√(0.55 × 0.45 / 64)

The expression  to find the 90% confidence interval is

\(CI = 0.55 + 1.645 \sqrt{\frac{0.55 ( 0.45)}{64} }\)

A confidence interval is a range of estimates for an unknown parameter. A confidence interval is computed at a designated confidence level; the 95% confidence level is most common, but other levels, such as 90% or 99%, are sometimes used.

CI             z

0.70        1.04

0.75 1.15

0.80 1.28

0.85 1.44

0.90        1.645

0.92 1.75

0.95 1.96

0.96 2.05

0.98 2.33

0.99 2.58

According to the question

Data of  people = 64

Data proportion (P) =  0.55

The 90% confidence interval

therefore, \(z_{c}\) =  1.645

CI = P± \(z \sqrt{\frac{P ( 1-P}{N} }\)

substituting the value

\(CI = 0.55 + 1.645 \sqrt{\frac{0.55 ( 0.45)}{64} }\)

CI = 0.5563

Hence, expression  to find the 90% confidence interval is

\(CI = 0.55 + 1.645 \sqrt{\frac{0.55 ( 0.45)}{64} }\)

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