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The monthly lease rates for a random sample of 8 new cars have a standard deviation of 41.5. Find the upper limit of the 90% confidence interval for the population standard deviation. Round your answer to the nearest tenth.

The monthly lease rates for a random sample of 8 new cars have a standard deviation-example-1
User Cruizh
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Answer

Upper Limit of the 90% confidence interval for the population standard deviation = 69.3

90% Confidence interval = 13.7 < σ < 69.3

Step-by-step explanation

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

Mathematically,

Confidence Interval = (Sample standard deviation) ± (Margin of error)

Sample standard deviation = 41.5

Margin of Error is the width of the confidence interval about the standard deviation.

It is given mathematically as,

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

Critical value at 90% confidence interval for sample size of 8 is obtained from the t-tables because the population standard deviation is unknown.

Degree of freedom = Sample size - 1 = n - 1 = 8 - 1 = 7

Significance level = 100 - (Confidence level) = 100 - 90 = 10% = 0.1

Critical value for degree of freedom 7, 0.10 significance level = 1.8946 (From the t-value calculator or the t-value tables)

Standard error of the distribution = σₓ = (σ/√n)

σ = standard deviation of the sample = 41.5

n = sample size = 8

σₓ = (41.5/√8) = 14.67

Confidence Interval = (Sample standard deviation) ± (Margin of error)

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

CI = 41.5 ± (1.8946 × 14.67)

CI = 41.5 ± 27.798

90% CI = (13.70, 69.30)

90% Confidence interval = 13.7 < σ < 69.3

Hope this Helps!!!

User Christoph Engwer
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