Question

The birth weights of full term babies born in Sydney are normally distributed. The management of a Sydney hospital is considering the resources needed to care for low birth-weight babies, and to this end, an analyst is doing some preliminary research on the distribution of birth-weights.

a. The analyst obtained a random sample of the weights of 51 full term babies recently born in Sydney. The sample mean was 2.98 kg and the sample standard deviation was 0.39 kg. Calculate a 90% confidence interval for mean baby weight and interpret your results.

b. Suppose it is known that the value of 0.39 is in fact, the population standard deviation. Re-calculate the 90% confidence interval for mean baby weight using this new information. Interpret your results and comment on any differences when compared to the interval calculated in part (a).

c. Would a 95% confidence interval be wider or narrower than the intervals obtained above? Explain briefly.

d. Assume that the analyst obtains a much larger sample (say, n=1,000) and recalculates the 95% confidence interval; will the width of the interval change? Explain your answer

e. In order to help the management of the hospital plan for the resources needed, the analyst also looks at the number of babies born with a low-birth weight, i.e. less than 2.5 kgs. The analyst was surprised to find that 6 of the babies in the sample have a low birth-weight. Using this information, calculate a 95% confidence interval for the proportion of low birth-weight babies.

Answer 1

A

Sample mean = 2.98 kg

Sample std dev = 0.39 kg

Standard error = Sample std dev / sqrt (Sample size)

= 0.39 / sqrt (51)

= 0.054

degrees of freedom (df) = sample size - 1

= 50

significance level (alpha) = 0.1

t value for given alpha and df = 1.67

Margin of error = t value * standard error

= 1.67 * 0.054

= 0.09

90% confidence interval = mean +/- margin of error

= 2.98 +/- 0.091

= 2.889 to 3.071

B

The only difference in calculation the Confidence Interval when we know the population standard deviation is to replace the t statistic with the z statistic which does not include degrees of freedom (or sample size) in calculationg the statistic.

So, the margin of error becomes -

z score * standard error

= 1.64 * 0.054

= 0.089

And confidence interval becomes -

2.891 to 3.069

C

A 95% confidence interval will be wider than the 90% confidence interval because the test statistic (z or t) will have a higher value which in turn will increase the margin of error and thereby confidence interval.

Intuitively, if you want to increase the chances of a sample mean fitting in the interval, you need to incraese it.

D

For a large sample size, the mean of the sample very much converges to the population mean and there is very less standard error (Which is the error in difference of means of different samples) and therefore, the margin of error becomes less and the width decreases

E

favourable cases = 6

proportion of babies having a low weight = p = 6/51

= 0.12

proportion of babies not having a low weight = p = 45/51

= 0.88

t score for a 0.05 alpha and 50 df = 2

Standard deviation = sqrt [ (p * q) / sample size]

= 0.046

Margin of error = 2* 0.046

= 0.092

Confidence interval

= 0.12 +/- 0.092

= 0.038 to 0.212