Question

John measured the total playing time of 30 randomly chosen CDs from his very large collection and found a mean of 77.9 minutes and a standard deviation of 11.6 minutes. Give the 93% confidence interval for the population mean, assuming that the population is approximately normally distributed. Round the endpoints to two decimal places.

Confidence interval:

tudies have suggusted that twins, in their early years, tend to have lower IQs and pick up language more slowly than non-twins. The slower intellectual growth may be caused by benign parental neglect. Suppose it is desired to estimate the mean attention time given to twins per week by their parents. A sample of 34 sets of two-year-old twin boys was taken, and after one week the attention time received was recorded. The mean was found to be 32.9 hours with a standard deviation of 14.3 hours. Use this information to contruct a 98% confidence interval for the mean attention time given to all twin boys by their parents, assuming that the population is approximately normally distributed. Round the endpoints to one decimal place.

Confidence interval:

Answer 1

Ansa:

Degrees of freedom = n - 1 = 30 - 1 = 29

For 30 degrees of freedom and 93% confidence, critical t value = 1.881

Hence,

93% confidence interval will be:

(tc*s/)

(77.91.881*11.6/)

77.93.984

(73.9,81.9)

Ansb:

Degrees of freedom = n - 1 = 34 - 1 = 33

For 33 degrees of freedom and 98% confidence, critical t value = 2.462

Hence,

98% confidence interval will be:

(tc*s/)

(32.92.445*14.3/)

32.95.996

(26.9,38.9)