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. A random sample of 100 children with working mothers showed that they were absent from school an average of 6 days per term with a standard deviation of 1.6 days. Provide a 98% confidence interval for the average number of days absent per term for all the children. 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) b. Determine a 98% confidence interval estimate for the average number of days absent per term for all the children. (10 points)

Answer :

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)

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