Mr. Crandall has assigned a term paper due at the end of the semester. He would like to know the average length of the papers. The data below are the he numbers of types pages from a random sample of 10 term papers. Use these data to compute a 95% confidence interval for the population mean length of all term papers for his class. Interpret the interval in the context of this problem. 14, 20, 25, 10, 16, 8, 15, 12, 18, 9. State what assumptions were met to use the method you selected. Check for the normal distribution of the population.

Find the mean first = sum of scores/number of scores

Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.

Standard deviation = square root of variance

95% = mean ± 1.96 SD

I'll let you do the calculations.

(16.66,12.74)

To compute a 95% confidence interval for the population mean length of all term papers, we can use the t-distribution since the sample size is small (<30) and the population standard deviation is unknown.

Step 1: Calculate the sample mean:
- Add up all the values: 14 + 20 + 25 + 10 + 16 + 8 + 15 + 12 + 18 + 9 = 147
- Divide by the sample size: 147 / 10 = 14.7

Step 2: Calculate the sample standard deviation:
- Subtract the sample mean from each value and square the result:
(14 - 14.7)^2, (20 - 14.7)^2, (25 - 14.7)^2, (10 - 14.7)^2, (16 - 14.7)^2, (8 - 14.7)^2, (15 - 14.7)^2, (12 - 14.7)^2, (18 - 14.7)^2, (9 - 14.7)^2
- Add up all these squared differences: 36.1 + 21.61 + 113.29 + 21.61 + 1.69 + 40.89 + 0.09 + 7.29 + 10.89 + 27.09 = 280.47
- Divide by (n-1) (where n is the sample size): 280.47 / 9 = 31.16
- Take the square root of the result: sqrt(31.16) ≈ 5.58

Step 3: Calculate the standard error:
- Divide the sample standard deviation by the square root of the sample size: 5.58 / sqrt(10) ≈ 1.76

Step 4: Find the critical value from the t-distribution:
- The degrees of freedom will be (n-1), where n is the sample size: (10 - 1) = 9
- Calculate the critical value using a t-table or software. For 95% confidence level and 9 degrees of freedom, the critical value is approximately 2.262.

Step 5: Calculate the margin of error:
- Multiply the standard error by the critical value: 1.76 * 2.262 ≈ 3.98

Step 6: Calculate the confidence interval:
- Subtract the margin of error from the sample mean: 14.7 - 3.98 ≈ 10.72
- Add the margin of error to the sample mean: 14.7 + 3.98 ≈ 18.68

The 95% confidence interval for the population mean length of all term papers is approximately 10.72 to 18.68 pages.

Interpretation: Based on the sample of 10 term papers, we are 95% confident that the population mean length of all term papers for Mr. Crandall's class falls between 10.72 and 18.68 pages. This means if we were to repeat this sampling process many times, 95% of the resulting confidence intervals would contain the true population mean length.

Assumptions: In order to use the t-distribution and perform a confidence interval analysis, we assume that the sample is random, the sample size is small enough, the population standard deviation is unknown, and the variable being measured (length of term papers) is approximately normally distributed in the population.

To compute a 95% confidence interval for the population mean length of all term papers, we can use the following steps:

Step 1: Calculate the sample mean (x̄) of the data set. Add up all the data values and divide by the sample size (n).

x̄ = (14 + 20 + 25 + 10 + 16 + 8 + 15 + 12 + 18 + 9) / 10 = 147 / 10 = 14.7

Step 2: Calculate the sample standard deviation (s) of the data set. This is a measure of how spread out the data is from the mean.

First, calculate the squared deviations from the mean for each data point:

(14 - 14.7)^2 = 0.49
(20 - 14.7)^2 = 27.04
(25 - 14.7)^2 = 109.29
(10 - 14.7)^2 = 21.16
(16 - 14.7)^2 = 1.69
(8 - 14.7)^2 = 44.89
(15 - 14.7)^2 = 0.09
(12 - 14.7)^2 = 7.29
(18 - 14.7)^2 = 10.89
(9 - 14.7)^2 = 32.49

Next, calculate the sum of these squared deviation values and divide by (n-1), then take the square root to find the sample standard deviation:

s = √((0.49 + 27.04 + 109.29 + 21.16 + 1.69 + 44.89 + 0.09 + 7.29 + 10.89 + 32.49) / (10-1))
s ≈ √(255.12 / 9) ≈ √28.35 ≈ 5.33

Step 3: Determine the sample size (n), which is given as 10 in the problem.

Step 4: Calculate the critical value for a 95% confidence level. Since we have a relatively small sample size (n<30) and the population is assumed to be normally distributed (more on this in the assumptions), we can use a t-distribution instead of the normal distribution. For a 95% confidence level with 9 degrees of freedom (n-1), the t-critical value is approximately 2.262.

Step 5: Calculate the margin of error (E) using the formula:

E = t * (s / √n)

E = 2.262 * (5.33 / √10) ≈ 3.57

Step 6: Calculate the lower and upper bounds of the confidence interval:

Lower bound = x̄ - E = 14.7 - 3.57 ≈ 11.13
Upper bound = x̄ + E = 14.7 + 3.57 ≈ 18.27

Therefore, the 95% confidence interval for the mean length of all term papers is roughly 11.13 to 18.27 pages. This means that we are 95% confident that the true population mean length of all papers falls within this interval.

Assumptions:
1. The sample is a random sample from the population of all term papers.
2. The term paper lengths are measured on a continuous scale.
3. The data are independent of each other.
4. The population of term paper lengths is approximately normally distributed. To check this assumption, we can construct a histogram or a probability plot of the data and visually inspect for a bell-shaped curve. Alternatively, we can perform a normality test such as the Shapiro-Wilk test or the Anderson-Darling test. If the assumption of normality is violated, alternative methods like bootstrapping or non-parametric tests may be considered.