Question

A company wishes to estimate the average amount the customers pay per month. A random sample of 200 customers' bills produced a sample mean of $50. Suppose that the population standard deviation is known and equals $5.

a) Find a 95% confidence interval for the population mean payment ( u ). The answers below may present approximate values.

options:

[49.307, 50.693]

[49.022, 50.983]

[3, 50]

[55.021, 80.981]

Answer 1

The 95% confidence interval for the population mean payment is approximately [49.307, 50.693], calculated using the sample mean, the z-score for a 95% confidence level, the population standard deviation, and the sample size.

Step-by-step explanation:

The student is asking how to find a 95% confidence interval for the population mean payment, denoted as μ (mu), when the population standard deviation is known. To calculate the 95% confidence interval, we use the z-score associated with the 95% confidence level and the known population standard deviation. The z-score for a 95% confidence level is approximately 1.96.

The formula for the confidence interval is:

Confidence Interval = x ± (z * (σ/√n))

Where:

• x is the sample mean
• z is the z-score corresponding to the confidence level
• σ is the population standard deviation
• n is the sample size

Applying the formula with the given values in the question:
Confidence Interval = $50 ± (1.96 * ($5/√200))

Confidence Interval = $50 ± (1.96 * ($5/14.14))

Confidence Interval = $50 ± (1.96 * 0.3535)

Confidence Interval = $50 ± 0.693

The 95% confidence interval for the population mean payment is roughly [49.307, 50.693].