Question

1. You measure 24 textbooks' weights, and find they have a mean weight of 75 ounces. Assume the population standard deviation is 3.3 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight.

2. You measure 37 backpacks' weights, and find they have a mean weight of 45 ounces. Assume the population standard deviation is 10.1 ounces. Based on this, construct a 95% confidence interval for the true population mean backpack weight.
3. You measure 30 watermelons' weights, and find they have a mean weight of 37 ounces. Assume the population standard deviation is 4.1 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean watermelon weight.
4. A student was asked to find a 99% confidence interval for widget width using data from a random sample of size n = 16. Which of the following is a correct interpretation of the interval 11.8 < μ < 20.4?
A. There is a 99% chance that the mean of a sample of 16 widgets will be between 11.8 and 20.4.
B. The mean width of all widgets is between 11.8 and 20.4, 99% of the time. We know this is true because the mean of our sample is between 11.8 and 20.4.
C. With 99% confidence, the mean width of all widgets is between 11.8 and 20.4.
D. With 99% confidence, the mean width of a randomly selected widget will be between 11.8 and 20.4.
E. There is a 99% chance that the mean of the population is between 11.8 and 20.4.
5. For a confidence level of 90% with a sample size of 23, find the critical t value.

Answer 1

(73.845 ; 76.155) ;

(41.633 ; 48.367) ;

1.273 ;

C. With 99% confidence, the mean width of all widgets is between 11.8 and 20.4. ;

1.717

Explanation:

1.)

Given :

Mean, xbar = 75

Sample size, n = 24

Sample standard deviation, s = 3.3

α = 90%

Confidence interval = mean ± margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 90% ; df = 24 - 1 = 23

Tcritical = 1.714

Margin of Error = 1.714 * 3.3/√24 = 1.155

Confidence interval = 75 ± 1.155

Confidence interval = (73.845 ; 76.155)

2.)

Given :

Mean, xbar = 45

Sample size, n = 37

Sample standard deviation, s = 10.1

α = 95%

Confidence interval = mean ± margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 95% ; df = 37 - 1 = 36

Tcritical = 2.028

Margin of Error = 2.028 * 10.1/√37 = 3.367

Confidence interval = 45 ± 3.367

Confidence interval = (41.633 ; 48.367)

3.)

Given :

Mean, xbar = 37

Sample size, n = 30

Sample standard deviation, s = 4.1

α = 90%

Margin of Error = Tcritical * s/√n

Tcritical at 90% ; df = 30 - 1 = 29

Tcritical = 1.700

Margin of Error = 1.700 * 4.1/√30 = 1.273

5.)

Sample size, n = 23

Confidence level, = 90%

df = n - 1 ; 23 - 1 = 22

Tcritical(0.05, 22) = 1.717