From 10 customers, mean weight 4.5 oz, s = 0.25 oz. Assuming random sampling, which method should be used to estimate the population mean weight of all cups with a confidence interval?

From 10 customers, mean weight 4.5 oz, s = 0.25 oz. Assuming random sampling, which method should be used to estimate the population mean weight of all cups with a confidence interval?

• A. One-Sample Z-Interval
• B. Two-Sample T-Interval
• C. One-Sample T-Interval
• D. Paired T-Interval

Answer: (C)

When estimating a population mean from a single random sample and the population standard deviation is unknown, use a one-sample t-interval. The data give a sample mean of 4.5 oz, a sample standard deviation of 0.25 oz, and a sample size of 10. Because the sample size is small and sigma is unknown, the t-distribution is appropriate rather than the normal z-distribution. The standard error is s/√n = 0.25/√10 ≈ 0.079, and the confidence interval is 4.5 ± t_{9,α/2} × 0.079. For a typical 95% confidence level, t_{9,0.025} ≈ 2.262, yielding about 4.5 ± 0.18, or roughly (4.32, 4.68). The other interval types don’t fit: a two-sample interval compares two groups, a paired interval is for matched pairs, and a z-interval would require knowing the population standard deviation or a very large sample.

When estimating a population mean from a single random sample and the population standard deviation is unknown, use a one-sample t-interval. The data give a sample mean of 4.5 oz, a sample standard deviation of 0.25 oz, and a sample size of 10. Because the sample size is small and sigma is unknown, the t-distribution is appropriate rather than the normal z-distribution. The standard error is s/√n = 0.25/√10 ≈ 0.079, and the confidence interval is 4.5 ± t_{9,α/2} × 0.079. For a typical 95% confidence level, t_{9,0.025} ≈ 2.262, yielding about 4.5 ± 0.18, or roughly (4.32, 4.68). The other interval types don’t fit: a two-sample interval compares two groups, a paired interval is for matched pairs, and a z-interval would require knowing the population standard deviation or a very large sample.