Question

The human resources department of a very large organisation is trying to determine the proportion of all employees that are satisfied with their current position. They randomly select 121 employees and ask them: "Are you satisfied with your current position?" 61 replied yes they were. Construct a 95% confidence interval to estimate the true proportion of all employees at this workplace who were satisfied with their position. A 95% confidence interval for the true proportion of all employees at this workplace who were satisfied with their position is between (round your answers to 2 dp) Answer and Answer.

Answer 1

Solution :

The 95% confidence interval to estimate the true population proportion is given as follows:

Where, p̂ is sample proportion, q̂ = 1 - p̂, n is sample size and Z(0.05/2) is critical z-value to construct 95% confidence interval.

Sample proportion of the employees who were satisfied with their current position is, p̂ = 61/121 = 0.5041

q̂ = 1 - 0.5041 = 0.4959 , n = 121

Using Z-table we get, Z(0.05/2) = 1.96

Hence, 95% confidence interval for the true proportion of all employees at this workplace who were satisfied with their position is,

On rounding to two decimal places we get,

Hence, a 95% confidence interval for the true proportion of all employees at this workplace who were satisfied with their position is between 0.42 and 0.59.

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