This is my first time taking statistics, please help me with the problem step by step.

A sample of 20 pages was taken without replacement from the 1,591-page phone directory
Ameritech Pages Plus Yellow Pages. On each page, the mean area devoted to display ads was measured
(a display ad is a large block of multicolored illustrations, maps, and text). The data (in
square millimeters) are shown below:

0 260 356 403 536 0 268 369 428 536
268 396 469 536 162 338 403 536 536 130

(a) Construct a 95 percent confidence interval for the true mean. (b) Why might normality be an
issue here? (c) What sample size would be needed to obtain an error of ±10 square millimeters
with 99 percent confidence? (d) If this is not a reasonable requirement, suggest one that is.

26% t(9)/h*4

Although I am not going to answer your questions, I want you to know that the area of the display ad on any page is not a "mean area."

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 (SD) = square root of variance

95% conf. interval = mean ± 1.96 SD

Is there a minimum requirement for the size of the ad on any page? If so, the distribution is unlikely to be normal.

I hope this helps a little.

Also, I don't see any connection of Lolli's post with your question..

To construct a confidence interval for the true mean, you can follow these steps:

Step 1: Calculate the sample mean (x̄)
First, add up all the values in your sample and divide it by the sample size (20) to find the sample mean. In this case:
x̄ = (0 + 260 + 356 + 403 + 536 + 0 + 268 + 369 + 428 + 536 + 268 + 396 + 469 + 536 + 162 + 338 + 403 + 536 + 536 + 130) / 20
x̄ ≈ 363.85

Step 2: Calculate the sample standard deviation (s)
Next, calculate the sample standard deviation to estimate the variability within the sample. You can use the standard deviation formula, but since the sample size is small and we don't have the population standard deviation, we'll use the unbiased estimator formula for the sample standard deviation. In this case:
s = sqrt([(Σx^2) - (n * x̄^2)] / (n - 1))
s = sqrt([∑(x^2) - (20 * 363.85^2)] / 19)
s ≈ 170.53

Step 3: Determine the critical value for the desired confidence level
To construct a 95% confidence interval, you need to determine the critical value. Since the sample size is small (n < 30) and the population standard deviation is unknown, you'll use the t-distribution instead of the standard normal distribution. You can use a t-table or a calculator to find the critical value. For a 95% confidence level and 19 degrees of freedom (20 - 1), the critical value is approximately 2.093.

Step 4: Calculate the margin of error (E)
The margin of error represents the maximum distance between the sample mean and the true population mean. To calculate the margin of error, use the formula:
E = (critical value) * (standard deviation / sqrt(sample size))
E = 2.093 * (170.53 / sqrt(20))
E ≈ 133.78

Step 5: Compute the confidence interval
Finally, you can construct the confidence interval by subtracting and adding the margin of error to the sample mean. The confidence interval formula is:
Confidence Interval = (sample mean - margin of error, sample mean + margin of error)
Confidence Interval ≈ (363.85 - 133.78, 363.85 + 133.78)
Confidence Interval ≈ (230.07, 497.63)

So, the 95 percent confidence interval for the true mean is (230.07, 497.63).

(b) Normality might be an issue here because in order for the confidence interval calculation to be valid, the underlying data should follow the normal distribution. If the data does not follow a normal distribution, the accuracy of the confidence interval may be compromised. In this case, the sample size is relatively small, so it is harder to assess whether the data follows a normal distribution.

(c) To determine the sample size needed to obtain an error of ±10 square millimeters with 99 percent confidence, you can use the formula:
n = [(Z * σ) / E]^2,
where n is the required sample size, Z is the standard score corresponding to the desired level of confidence (in this case, the confidence level is 99%, so Z ≈ 2.576), σ is the standard deviation (which we estimated as s ≈ 170.53), and E is the desired error limit (±10 square millimeters).

Plugging in the values, we get:
n = [(2.576 * 170.53) / 10]^2
n ≈ 1139.23
Since the sample size must be an integer, you would round up to the nearest whole number. Therefore, the sample size needed to obtain an error of ±10 square millimeters with 99 percent confidence is 1140.

(d) If obtaining a sample size of 1140 is not feasible for practical reasons, you may consider reducing the desired level of confidence. Lowering the confidence level will result in a smaller sample size requirement. For example, if you decrease the confidence level to 95%, you can recalculate the sample size using the same formula and obtain a smaller value.