Question

IMR in Singapore. The infant mortality rate (IMR) is the number of infant deaths per 1000 live births. Suppose that you have been commissioned to estimate the IMR in Singapore. From a random sample of $1109$live births in Singapore, you find that $0.361\%$of them resulted in infant deaths. You next find a $90\%$confidence interval:

$0.00361\pm 1.645·\sqrt{(0.00361)(0.99639)/1109}\text{,}$

or $0.000647$to $0.00657$. You then conclude, "I can be $90\%$confident that the IMR in Singapore is somewhere between $0.647$and $6.57$." How did you do?

Short answer

It is incorrect to calculate IMR with the given sample data using the given approach.

Step-by-step solution

Given Information

Sample size is $n=1109$.

Sample proportion is $\hat{p}=0.361\%$

$=0.00361$.

Explanation

The $90\%$confidence interval is as follows:

A confidence interval of $1-\alpha$,

To find $z_{\alpha /2}$using table II.

$\hat{p}\pm z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$\Rightarrow \text{Confidence interval}=0.00361\pm 1.645\sqrt{\frac{0.00361(1-0.00361)}{1109}}$

$\Rightarrow \mathrm{CI}=0.00361\pm 0.00296$

$\Rightarrow \mathrm{CI}=0.000647\text{to}0.00657$$\Rightarrow \mathrm{CI}=0.0647\%\text{to}0.657\%$

One can be $90\%$confident that infant mortality rate(IMR) in Singapore is somewhere between $0.0647\%$to $0.657\%$.

Explanation

To find the confidence interval in the present scenario.

The number of samples is $n=1109$.

The ratio of the sample is $\hat{p}=0.361\%$

$=0.00361$

The number of success $x=n\times \hat{p}$

$=1109\times 0.361\%$

$=4$

(less than $5$).

Since assumption that localid="1651310401339" $(x>5)$is not being satisfied here.