Answer:
a) The equation is:
Confidence Interval = Mean ± Z score × Standard deviation/√Number of samples
b) The 98% confidence interval = (5.62784, 6.37216)
Step-by-step explanation:
a. Write down the equation you should use to construct the confidence interval for the average number of days absent per term for all the children. (10 points)
Confidence Interval = Mean ± Z score × Standard deviation/√Number of samples
b. Determine a 98% confidence interval estimate for the average number of days absent per term for all the children. (10 points)
Confidence Interval = Mean ± Z score × Standard deviation/√Number of samples
Mean = 6 days
Standard deviation = 1.6 days
Number of samples = 100
Z score of 98% confidence interval = 2.326
Confidence interval = 6 ± 2.326 × 1.6/√100
= 6 ± 2.326 × 1.6/10
= 6 ± 0.37216
= 6 - 0.37216
= 5.62784
6 + 0.37216
= 6.37216
Therefore, the 98% confidence interval = (5.62784, 6.37216)
12/14=6/7
Because you divide by 2.
You have to divide 108 with 25.18
about 4.3/1
Answer:
x is 3.4 i think
Step-by-step explanation:
