When establishing the confidence interval for the average weight of a cereal box,assume that the population standard deviation is known to be 2 ounces.Based on a sample,the average weight of a sample of 20 boxes is 16 ounces.The appropriate test statistic to use is ________.
A) t
B) z
C) χ
D) p

Question

Quizplus · Accepted Answer

Since the population standard deviation is known and the sample size is larger than 30, we can use a z-test. The formula for the z-score in this scenario is:

z = (sample mean - population mean) / (population standard deviation / sqrt(sample size))

Plugging in the values given:

z = (16 - population mean) / (2 / sqrt(20))

Since we are trying to find a confidence interval, we need to find the critical value (z*) from the standard normal distribution table. Assuming a 95% confidence level (alpha = 0.05), we need to find the z-value that corresponds to a tail area of 0.025 (0.5 - 0.025 = 0.475 on each side).

Using a standard normal distribution table, we find that z* = 1.96.

Now we can find the margin of error:

Margin of error = z* * (population standard deviation / sqrt(sample size))

Margin of error = 1.96 * (2 / sqrt(20)) = 0.872

Finally, we can construct the confidence interval:

Sample mean +/- margin of error

16 +/- 0.872

CI = (15.13, 16.87)

Therefore, we can say with 95% confidence that the true average weight of a cereal box is between 15.13 and 16.87 ounces.

When Establishing the Confidence Interval for the Average Weight of a Cereal