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)
Answer:
36 because the absolute value of -9-3 is 12, and then 12 times 3 is 36
Answer:
75% of 124 is 93.
Step-by-step explanation:
I hoped this helped ^^
Answer:
B
Step-by-step explanation:
The format for quadrant 2 would be (-,+) and answer choice be fits that because it is(-3,2)
