Question

0 3 1 20 9 5 10 1 10 4 14 2 4 4 5 A random sample of American children was taken to estimate the total number of hours they spend watching cartoons in an average week. The responses above were chosen from a simple random sample of 15 children and the data was normally distributed. Create a 98% confidence interval for the mean number of hours American children spend watching cartoons in one week. 1st: Identify point estimate                                            2nd: Requirements 3rd: Find critical value (include sketch!)                     4th: Find margin of error 5th: Construct confidence interval                               6th: Interpretation

Answer 1

Answer 3: A random sample of American children was taken to estimate the total number of hours they spend watching cartoons in an average week. The responses above were chosen from a simple random sample of 15 children and the data was normally distributed.

Create a 98% confidence interval for the mean number of hours American children spend watching cartoons in one week.

Solution:

1) point estimate = x̄

Mean, x̄ = Σx/n

Mean, x̄ = 92/15

Mean, x̄ = 6.1333

Std.Dev, s = √Σ(xi - x̄)^2/n-1

s = √Σ(xi - 6.1333)^2/15-1

s = √425.7333 / 14

Std.Dev, s = 5.5145

2) Requirements:

Random sample of 15 students.

Sample size is less than 30 we use t distribution.

3) Critical value:

At 98% confidence interval, α = 0.02

df = n-1 = 15 - 1 = 14

t critical = t(α/2,df) = t(0.01,14)

t critical = 2.6245

4) Margin of error, E:

E = t critical * s / √n

E = 2.6245 * 5.5145 / √15

E = 3.7369

5) the 98% confidence interval for mean number of hours:

CI = x̄ ± E

CI = 6.1333 ± 3.7369

CI = (2.3964, 9.8702)

6) Interpretation:

Therefore, the 98% confidence interval for the mean number of hours American children spend watching cartoons in one week is between 2.3964 and 9.8702.