A national survey of companies included a question that asked whether the company had at least one bilingual telephone operator.

Question

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A national survey of companies included a question that asked whether the company had at least one bilingual telephone operator.

The sample results of 90 companies (Y denotes that the company does have at least one bilingual operator; N denotes that it does not) are listed in the Excel file Bilingual.xlsx. Use the information to estimate with 80% confidence the proportion of the population that does have at least one bilingual operator. Round LCL and UCL values to 3 decimal places. Data file: Bilingual.xlsxPreview the document Group of answer choices LCL = 0.270, and UCL = 0.397 LCL = 0.252, and UCL = 0.415 LCL = 0.236, and UCL = 0.431 LCL = 0.205, and UCL = 0.461

Mathematics

1 answer:

nirvana33 [79] · Answer 1 · 2022-05-06T00:19:33+03:00

Answer:

The first option is correct. Option A is correct.

LCL = 0.270, and UCL = 0.397

80% Confidence interval = (0.270, 0.397)

Step-by-step explanation:

The data for Y and N for the 90 companies is attached to this solution provided.

Y represents companies with at least 1 bilingual operator and N represents companies with no bilingual operator.

The number of Y in the data = 30

Hence, sample proportion of companies with at least one bilingual operator = (30/90) = 0.3333

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 = 0.3333

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 80% confidence level for sample size of 90 is obtained from the z-tables.

Critical value = 1.280

Standard error of the mean = σₓ = √[p(1-p)/n]

p = sample proportion

n = sample size = 90

σₓ = √[0.3333×0.6667)/90] = 0.0496891568 = 0.04969

80% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.3333 ± (1.28 × 0.04969)

CI = 0.3333 ± 0.0636021207

80% CI = (0.2696978793, 0.3969021207)

80% Confidence interval = (0.270, 0.397)

Hope this Helps!!!