Question

Solve for 1 and 2:

1) Suppose we were to gather a random sample of 15 observations from a population and wished to calculate an 80% confidence interval for the mean, µ, in the case where the population standard deviation, σ, is unknown. Enter the value from the Student's t distribution that we would use, to three decimal places.

2)A random sample of 51 undergraduate statistics students resulted in a sample mean age of 22.1 years, with a sample standard deviation of 2.9 years. Find the upper bound of the 99% confidence interval for the true mean age, to one decimal place.

Answer 1

Answer 2

1. The value from the Student's t distribution that we would use for a random sample of 15 observations and an 80% confidence interval is 1.753 (rounded to three decimal places).

2. The upper bound of the 99% confidence interval for the true mean age can be calculated using the formula:

Upper bound = sample mean + (t-value)*(standard error)

where the t-value is found using the t-distribution with degrees of freedom n-1 (where n is the sample size) and the standard error is calculated as the sample standard deviation divided by the square root of the sample size.

Plugging in the given values, we have:

t-value for 99% confidence interval with 50 degrees of freedom = 2.680 (using a t-distribution table or calculator) standard error = 2.9 / sqrt(51) = 0.406 sample mean = 22.1

Therefore, the upper bound of the 99% confidence interval for the true mean age is:

Upper bound = 22.1 + (2.680)*(0.406) = 23.2 (rounded to one decimal place)