- #1

ianrice

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59, 84, 68, 93, 49, 77, 82, 75, 81, 58, 70, 80

Find the 90% two-sided confidence interval for the mean score.

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- Thread starter ianrice
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In summary, the conversation discusses the distribution of grades and the unknown mean and variance. A random sample of 12 students is given and the task is to find the 90% confidence interval for the mean score. It is also mentioned that the grading system may not allow for negative grades, which would affect the normal distribution assumption. The use of a Student's t-distribution is suggested as an alternative.

- #1

ianrice

- 7

- 0

59, 84, 68, 93, 49, 77, 82, 75, 81, 58, 70, 80

Find the 90% two-sided confidence interval for the mean score.

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- #2

Stephen Tashi

Science Advisor

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ianrice said:The distribution of grades is know to follow a normal distribution

Does this grading system allow negative grades? If not, the the grades can't actually be normally distributed.

Of course, perhaps this is a homework problem where such technicalities are ignored. If so, you should post it somewhere in the "Homework & Coursework Questions" section of the forum and show your attempt at solution.

- #3

statdad

Homework Helper

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Nothing is ever really normally distributed: if the standard deviation is small enough then to a very good approximation the normal distribution can be quite useful.

- #4

DrDu

Science Advisor

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You might consider a Student's t-distribution

- #5

blue_raver22

- 2,250

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Based on the given information, we can use the t-distribution to find the 90% two-sided confidence interval for the mean score. First, we calculate the sample mean and standard deviation using the given scores:

Sample mean, x̄ = (59+84+68+93+49+77+82+75+81+58+70+80)/12 = 73.5

Sample standard deviation, s = √[(1/(12-1)) * ((59-73.5)^2 + (84-73.5)^2 + (68-73.5)^2 + (93-73.5)^2 + (49-73.5)^2 + (77-73.5)^2 + (82-73.5)^2 + (75-73.5)^2 + (81-73.5)^2 + (58-73.5)^2 + (70-73.5)^2 + (80-73.5)^2)] = 13.9

Next, we need to determine the critical value for a 90% confidence interval with 12-1 = 11 degrees of freedom. Using a t-table, we find the critical value to be 1.796.

Finally, we can calculate the confidence interval using the formula:

x̄ ± tα/2 * (s/√n)

Where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and tα/2 is the critical value.

Plugging in the values, we get:

73.5 ± 1.796 * (13.9/√12) = (67.4, 79.6)

Therefore, we can be 90% confident that the true mean score for all students falls within the interval of 67.4 to 79.6. This means that if we were to take multiple random samples of 12 students and calculate the confidence interval each time, 90% of those intervals would contain the true mean score.

A 90% two-sided confidence interval is a range of values that is computed from a sample of data and is likely to contain the true population mean with 90% confidence. This means that if the same sample is taken multiple times, 90% of the resulting confidence intervals will contain the true population mean.

A 90% two-sided confidence interval is calculated using the sample mean, standard deviation, and sample size. The formula for a 90% two-sided confidence interval is: sample mean ± (1.645 * standard deviation / square root of sample size).

A 90% confidence level is commonly used because it strikes a balance between being too narrow and too wide. A narrower confidence interval (e.g. 95% or 99%) may not capture the true population mean as often, while a wider interval (e.g. 80%) may be too conservative and capture the true mean too often.

The 90% confidence interval tells us that there is a 90% probability that the true population mean falls within the calculated range. It also provides an estimate of the precision of the sample mean by indicating the range within which the true mean is likely to fall.

The sample size has an inverse relationship with the width of the 90% confidence interval. As the sample size increases, the standard error decreases, resulting in a narrower confidence interval. This is because a larger sample size provides more information and reduces the margin of error in the estimation of the true population mean.

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