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The Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 16 people reveals the mean yearly consumption to be 27 kg with a sample standard deviation of 9 kg. Assume a normal population. Develop a 90% confidence interval for the mean annual consumption of sugar. Round final answers to two decimal places

User Zdebruine
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Final answer:

To calculate the 90% confidence interval for the mean annual sugar consumption, we use the t-distribution with a sample size of 16, sample mean of 27 kg and standard deviation of 9 kg, which results in a confidence interval of approximately 23.06 kg to 30.94 kg.

Step-by-step explanation:

The question requires us to calculate a 90% confidence interval for the mean annual consumption of sugar using a sample mean (X-bar) of 27 kg, sample standard deviation (s) of 9 kg, and a sample size (n) of 16. Since we are dealing with the sample size less than 30 and the population standard deviation is unknown, we should use the t-distribution. The degrees of freedom (df) for this estimation will be n - 1 which is 15. The t-score for a 90% confidence level and df of 15 can be found using a t-distribution table or a calculator and is approximately 1.753.

To compute the margin of error (E), we use the formula: E = t-score * (s / √n). Lastly, the confidence interval is given by: (X-bar - E, X-bar + E).

Step-by-step calculation:

  1. Calculate the degrees of freedom: df = n - 1 = 15.
  2. Identify the t-score for a 90% confidence interval and df of 15, which is approximately 1.753.
  3. Compute the margin of error (E): E = 1.753 * (9 / √16) = 1.753 * (9 / 4) ≈ 3.94275.
  4. Find the confidence interval: (27 - 3.94, 27 + 3.94) which is (23.06, 30.94) after rounding to two decimal places.

Therefore, the 90% confidence interval for the mean annual sugar consumption is approximately between 23.06 kg and 30.94 kg.

User Tuan Anh Vu
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