Question

The Baltimore Sun reported on a study by Dr. Sara Harkness in which she compared the sleep patterns of 6-month-old infants in the United States and the Netherlands. She found that the 36 U.S. infants slept an average of just under 13 hours out of 24, whereas the 66 Dutch infants slept an average of almost 15 hours. Suppose it was 0.5 hours for each group.
a. Compute the standard error of the mean (SEM) for the U.S. babies.
b. Compute an approximate 95% confidence interval for the mean sleep time for 6-month-old babies in the United States.
c. Compute an approximate 95% for the difference in average sleep time for 6-month-old Dutch and U.S. infants.

Answer 1

(a) The standard error of the mean (SEM) for the U.S. babies is 0.083.

(b) An approximate 95% confidence interval for the mean sleep time for 6-month-old babies in the United States is 12.8367< μ1< 13.163.

(c) An approximate 95% for the difference in average sleep time for 6-month-old Dutch and U.S. infants is −2.203< μ1 <μ2 < −1.7969.

What is standard error and how it is measured?

A statistical concept known as the standard error uses standard deviation to assess how accurately a sample distribution represents a population. The standard error of the mean describes the statistical variation between a sample mean and the population's actual mean.

According to the given information:

In United States:

Sample size $\boldsymbol{n}_1=\mathbf{3 6}$

Mean, $\bar{x}_1=13$

Standard deviation, $s_1=\mathbf{0 . 5}$

In Netherlands:

Sample size, $n_2=66$

Mean, $\bar{x}_2=15$

Standard deviation, $s_2=\mathbf{0 . 5}$

(a) For the newborns in the United States, the standard error of the mean (SEM) is determined as follows:

$\begin{aligned}& S E\left(\bar{x}_1\right)=\frac{s_1}{\sqrt{n_1}} \\& S E\left(\bar{x}_1\right)=\frac{0.5}{\sqrt{36}} \\& S E\left(\bar{x}_1\right)=0.083\end{aligned}$

(b) The following formula is used to determine the 95% confidence interval for the mean sleep time for babies aged 6 months in the United States:

$\bar{x}_1 \pm z_{0.05 / 2} \times \sqrt{\frac{s_1^2}{n_1}}$

$\begin{aligned}& 13 \pm 1.96 \times \sqrt{\frac{0.5^2}{36}} \\& 12.8367 < \mu_1 < 13.163\end{aligned}$

(c) The 95% difference in the amount of time newborns in the Netherlands and the United States spend sleeping on average is computed as follows:

$\begin{aligned}& \bar{x}_1-\bar{x}_2 \pm z_{0.05 / 2} \times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}} \\& (13-15) \pm 1.96 \times \sqrt{\frac{0.5^2}{36}+\frac{0.5^2}{66}} \\& -2.203 < \mu_1 < \mu_2 < -1.7969\end{aligned}$

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