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The mean of 11 determinations is 8.37 with a standard deviation of 0.17, and the true value is 7.19. Determine if there is a significant difference between the mean and the known/true value at 95% confidence.

a) There is a significant difference.
b) There is no significant difference.
c) Insufficient information to determine.
d) The confidence level is not specified.

User LeandroG
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1 Answer

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Final answer:

A 95% confidence interval is constructed to determine if there's a significant difference between the sample mean of 8.37 and the true value of 7.19. The interval calculated is (8.27, 8.47), which does not contain the true value of 7.19, hence there is a significant difference at the 95% confidence level.

Step-by-step explanation:

The question is asking whether there is a significant difference between the sample mean (8.37) of 11 determinations and the known true value (7.19) based on a 95% confidence interval. To determine this, we use the concept of a confidence interval for the mean.

A 95% confidence interval means that if we were to take many samples and create a confidence interval from each, we would expect 95% of those intervals to contain the true population mean. This interval can be constructed using the sample mean, the standard deviation, and the appropriate z-score or t-score for the confidence level and sample size.

To find this interval, we typically use the formula: Mean ± (z-score)(standard deviation)/sqrt(n), where n is the sample size. Knowing that the standard deviation is 0.17 and the sample size is 11, and using the z-score for a 95% confidence interval for a normal distribution (which is approximately 1.96), we can calculate the interval:

8.37 ± (1.96)(0.17)/sqrt(11) = (8.37 ± 0.10) = (8.27, 8.47).

Since the true value of 7.19 is not within this interval, we conclude that there is a significant difference between the sample mean and the true value at the 95% confidence level.

User Adolf
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