While writing an article on the high cost of college education, a reporter took a random sample of the cost of new textbooks for a semester. The random variable x is the cost of one book. Her sample data can be summarized by the following. (Give your answers correct to two decimal places.)
n = 21, Óx = 3635.6, and Ó(x - x)2 = 9750.4
(a) Find the sample mean, x.
$ 173.12.
(b) Find the sample standard deviation, s.
$ 22.08 .
(c) Find the 90% confidence interval to estimate the true mean textbook cost for the semester based on this sample.
Lower Limit $
Upper Limit $
I can not get the limits to work out I have tried 4 times and I only have one more time. I have plugged information into calculator that worked on other limits but not this one.
To find the 90% confidence interval, we first need to calculate the margin of error (E).
The formula for the margin of error is:
E = Z * (σ / √n)
Where:
Z is the z-value corresponding to the desired confidence level (90% confidence level is 1.645, which can be obtained from a z-value table),
σ is the sample standard deviation, and
n is the sample size.
Given:
n = 21,
σ = √(9750.4) ≈ 98.73.
Calculating the margin of error:
E = 1.645 * (98.73 / √21) ≈ 33.40.
Now, to find the confidence interval, we use the formula:
Confidence Interval = (x - E, x + E)
Given:
x = 173.12 (sample mean).
Calculating the confidence interval:
Lower Limit = 173.12 - 33.40 ≈ 139.72.
Upper Limit = 173.12 + 33.40 ≈ 206.52.
Therefore, the 90% confidence interval to estimate the true mean textbook cost for the semester based on this sample is:
Lower Limit: $139.72
Upper Limit: $206.52
To find the 90% confidence interval for estimating the true mean textbook cost for the semester, we can use the formula:
Confidence Interval = x ± (t * (s / √n))
where x is the sample mean, s is the sample standard deviation, n is the sample size, and t is the critical value for a 90% confidence level.
Let's calculate each component step by step:
(a) Sample mean:
Given that the sample size is n = 21, and the sum of the values is Σx = 3635.6, we can find the sample mean x by dividing the sum by the sample size:
x = Σx / n
x = 3635.6 / 21
x ≈ 173.12
Therefore, the sample mean is $173.12.
(b) Sample standard deviation:
Given that the sum of squared deviations from the mean is Σ(x - x)² = 9750.4, we can find the sample standard deviation s using the formula:
s = √(Σ(x - x)² / (n - 1))
s = √(9750.4 / (21 - 1))
s ≈ 22.08
Therefore, the sample standard deviation is $22.08.
(c) Confidence interval:
Since we are looking for a 90% confidence interval, we need to find the critical value for a 90% confidence level. The critical value can be found using a t-distribution with n - 1 degrees of freedom. In this case, with 21 - 1 = 20 degrees of freedom, the critical value is approximately 1.725.
Now, we can calculate the confidence interval using the formula mentioned earlier:
Confidence Interval = x ± (t * (s / √n))
Confidence Interval = 173.12 ± (1.725 * (22.08 / √21))
Calculating the expression in parentheses:
(1.725 * (22.08 / √21)) ≈ 7.99
Therefore, the confidence interval is approximately:
Lower Limit = 173.12 - 7.99 ≈ $165.13
Upper Limit = 173.12 + 7.99 ≈ $181.11
Hence, the 90% confidence interval to estimate the true mean textbook cost for the semester based on this sample is approximately $165.13 to $181.11.