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Butoxors [25]
3 years ago
7

. Construct the 90% confidence interval for the population proportion of failures. (Round intermediate calculations to at least

4 decimal places. Round "z" value and final answers to 3 decimal places.)
Mathematics
1 answer:
ollegr [7]3 years ago
6 0

Answer:

a) 90% Confidence interval For population proportion of successes = (0.315, 0.485)

b) 90% Confidence interval For population proportion of failures = (0.515, 0.685)

Step-by-step explanation:

A random sample of 90 observations results in 36 successes. [You may find it useful to reference the z table.]

a. Construct the a 90% confidence interval for the population proportion of successes. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)

Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion = proportion of success = (36/90) = 0.40

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 90% confidence interval

The question asks us to use the z-value. We should be using the t- value though as we do not have any information on the population standard deviation.

Critical value = 1.645 (from the z-tables)

Standard error of the mean

= σₓ = √[p(1-p)/n]

p = 0.40

n = sample size = 90

σₓ = √(0.4×0.6/90) = 0.0516

90% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.40 ± (1.645 × 0.0516)

CI = 0.40 ± 0.0849

90% CI = (0.315, 0.485)

90% Confidence interval For successes = (0.315, 0.485)

b. Construct the a 90% confidence interval for the population proportion of failures. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion = proportion of failures = 1 - 0.40 = 0.60

Margin of Error = (Critical value) × (standard Error)

Critical value at 90% confidence interval

The question asks us to use the z-value. We should be using the t- value though as we do not have any information on the population standard deviation.

Critical value = 1.645 (from the z-tables)

Standard error of the mean

= σₓ = √[p(1-p)/n]

p = 0.60

n = sample size = 90

σₓ = √(0.6×0.4/90) = 0.0516

CI = 0.60 ± (1.645 × 0.0516)

CI = 0.60 ± 0.0849

90% CI = (0.515, 0.685)

90% Confidence interval For failures = (0.515, 0.685)

Hope this Helps!!!

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