Demand per hour for gasoline at a local station is normally distributed with a mean of 1510 gallons and std deviation of 57 gall
1 answer:
You might be interested in
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!!!
