A class survey in a large class for first-year college students asked, "About how many minutes do you study on a typical weeknight?" The mean response of the 463 students was x = 118 minutes.
Suppose that we know that the study time follows a Normal distribution with standard deviation σ = 65 minutes in the population of all first-year students at this university.
Use the survey result to give a 99% confidence interval for the mean study time of all first-year students.
95% = mean ± 2.575 SEm
SEm = SD/√n
I'll let you do the calculations.
Suppose x has a distribution with μ = 11 and σ = 5.
(a) If a random sample of size n = 45 is drawn, find μx, σ x and P(11 ≤ x ≤ 13). (Round σx to two decimal places and the probability to four decimal places.)
It is 99%, sorry for the typo.
To find a confidence interval for the mean study time of all first-year students, we can use the formula for a confidence interval:
CI = x ± Z * (σ / √n)
Where:
- CI represents the confidence interval
- x is the sample mean
- Z is the critical value from the standard normal distribution that corresponds to the desired level of confidence
- σ is the population standard deviation
- n is the sample size
In this case, we know that x = 118 minutes, σ = 65 minutes, and n = 463 students.
To find the critical value Z for a 99% confidence level, we need to determine the area that is outside of the confidence interval. Since we want a 99% confidence level, the area outside of the interval is (1 - 0.99) / 2 = 0.005 on each tail.
We can look up this area in the standard normal distribution table and find that the critical value Z is approximately 2.576.
Now, let's plug in the values into the formula:
CI = 118 ± 2.576 * (65 / √463)
Calculating the values:
CI = 118 ± 2.576 * (65 / √463)
CI = 118 ± 2.576 * (65 / 21.52)
CI = 118 ± 2.576 * 3.021
CI = 118 ± 7.790
Therefore, the 99% confidence interval for the mean study time of all first-year students is (110.21, 125.79) minutes.