Question

1. A random sample of 49 children with working mothers showed that they were absent from school an average of 6 days per term with a standard deviation of 1.8 days.

1. 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. (6 points)

2. Determine a 98% confidence interval estimate for the average number of days absent per term for all the children. (8 points)

3. Determine a 95% confidence interval estimate for the average number of days absent per term for all the children. (8 points)

4. Discuss why 98% and 95% confidence intervals are different. (6 points)

Answer 1

Let µ denote the average number of days absent per term for all the children

According to data the sample mean x̅ = 6, sample standard deviation s = 1.8 and sample size n = 49

Let the level of significance be  α

Solution to a)

The equation is

Solution to b)

Here α = 2% = 0.02

So 98% Confidence Interval is

= (5.381165, 6.618835)

Hence 98% Confidence Interval for µ is (5.381165, 6.618835)

Solution to c)

Here α = 5% = 0.05

So 95% Confidence Interval is

= (5.48298, 6.51702)

Solution to d)

A 95% confidence interval is a range of values that you can be 95% certain contains the true mean of the population.

A 98% confidence interval is a range of values that you can be 98% certain contains the true mean of the population.

So that is why both are different and mathematically as the critical value for both differ so both are different.