A simple random sample of seven (7) students is selected, and the students are asked how much time they spent studying for their University subjects (combined) outside class times. From past experience, it is known that these times (in hours) are distributed normally.The times (in hours) for the sample of students are as follows:2.3 7.2 4.5 12.5 6.6 2.5 5.5Based on these results, a confidence interval estimate for the population mean is found to be 5.9 ± 2.6.The level of confidence the researcher has used is closest to:-Group of answer choices95%90%97.5%99%
Question
A simple random sample of seven (7) students is selected, and the students are asked how much time they spent studying for their University subjects (combined) outside class times. From past experience, it is known that these times (in hours) are distributed normally.The times (in hours) for the sample of students are as follows:2.3 7.2 4.5 12.5 6.6 2.5 5.5Based on these results, a confidence interval estimate for the population mean is found to be 5.9 ± 2.6.The level of confidence the researcher has used is closest to:-Group of answer choices95%90%97.5%99%
Solution 1
The correct answer is: 90%.
Here's why:
The confidence interval is calculated as the sample mean ± the margin of error. The margin of error is the product of the standard deviation (or standard error) and the z-score corresponding to the desired level of confidence.
In this case, the sample mean is 5.9 and the margin of error is 2.6. The z-score that corresponds to a 90% confidence level is approximately 1.645 (for a two-tailed test), which is less than the z-score for a 95% confidence level (approximately 1.96) and much less than the z-scores for 97.5% and 99% confidence levels.
Therefore, the level of confidence that would result in a margin of error of 2.6 (given the sample mean of 5.9 and the sample data provided) is closest to 90%.
Solution 2
The correct answer is: 95%
Here's why:
The margin of error in the confidence interval is 2.6. The sample mean is 5.9. Therefore, the confidence interval is (5.9 - 2.6, 5.9 + 2.6) = (3.3, 8.5).
The sample standard deviation can be calculated from the given data. The standard deviation of the sample is approximately 3.4.
The standard error of the mean is the standard deviation divided by the square root of the sample size, which is 3.4 / sqrt(7) = 1.28.
The z-score for the confidence interval is the margin of error divided by the standard error, which is 2.6 / 1.28 = 2.03.
Looking up this z-score in a standard normal distribution table, we find that the corresponding level of confidence is closest to 95%.
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