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ryzh [129]
3 years ago
9

To the following is not a network connection LAM MAN SAN WAN

Computers and Technology
1 answer:
Sedaia [141]3 years ago
3 0
SAN is not a network connection.

Can I have Brainlest please?
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The Baltimore Sun reported on a study by Dr. Sara Harkness in which she compared the sleep patterns of 6-month-old infants in th
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(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:

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

To know more about standard error:

brainly.com/question/29363195

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