Question

We took a random sample of 100 customers in a grocery store. These customers spent overall amount of $2,500 in the grocery store (total amount spent by a sample of 100 customers). Assume true variance of amount spent by a random customer is σ² =81 and answer the following questions: a. Find the variance of total amount spent by 100 customers (σ²ₜ). b. Construct a two-sided 95% confidence interval for μ (true average spent by a customer). Notice that an estimate for total amount spent by 100 customers is given ( T =$2,500), and you need to find the estimate of average amount spent by a typical customer.

Answer 1

The variance of the total amount spent by 100 customers is 8100. The two-sided 95% confidence interval for the true average amount spent by a customer is ($23.236, $26.764).

Step-by-step explanation:

To address this question, we first calculate the variance of the total amount spent by 100 customers. The variance of the total is the variance of a single customer multiplied by the sample size (n), hence σ²ₜ = n × σ², where n=100 and σ²=81. This gives us σ²ₜ = 8100.

To construct a two-sided 95% confidence interval for μ (true average spent by a customer), we use the formula for the confidence interval of the population mean when the population standard deviation is known: µ ≈ μ ± Z*(σ/√n), where μ is the sample mean, σ is the population standard deviation, Z* is the Z-score corresponding to the desired confidence level, and n is the sample size.

Since the sample mean (μ) can be calculated as the total amount spent by 100 customers divided by 100, we get μ = $2,500 / 100 = $25.

For a 95% confidence interval, we use a Z-score of approximately 1.96. The standard deviation (σ) is the square root of the variance, which is 9 (since σ² = 81). So, we calculate the margin of error (ME) as ME = 1.96*(9/√100) = 1.96*(9/10) = 1.764.

The confidence interval is therefore $25 ± $1.764. So, the interval is ($23.236, $26.764).