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 = 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.
(a) Use the survey result to give a 99% confidence interval for the mean study time of all first-year students.
(b) What condition not yet mentioned must be met for your confidence interval to be valid?
http://davidmlane.com/hyperstat/z_table.html
To calculate a 99% confidence interval for the mean study time of all first-year students, we need to use the sample mean and the standard deviation of the population.
(a) The formula for calculating the confidence interval is:
Confidence Interval = sample mean ± (z-value * standard deviation)
Since we want a 99% confidence interval, we need to find the critical z-value for a 99% confidence level. The z-value can be calculated using a standard normal distribution table or a calculator. For a 99% confidence level, the z-value is 2.576.
Now we can calculate the confidence interval using the given information:
Confidence Interval = 118 ± (2.576 * 65)
Confidence Interval = 118 ± 167.74
Confidence Interval = (−49.74, 285.74)
Therefore, the 99% confidence interval for the mean study time of all first-year students is (-49.74, 285.74) minutes.
(b) The condition that must be met for the confidence interval to be valid is that the sample data should be randomly selected and representative of the population. Random selection ensures that the sample is not biased towards a specific group or characteristic. Representativeness means that the sample accurately reflects the characteristics of the population. If the sample is not random or representative, the confidence interval may not provide accurate estimates of the population mean.