All heat engines involve combustion. True or False

Computers and Technology

2 answers:

lakkis [162]4 years ago

5 0

The correct answer is False

exis [7]4 years ago

4 0

The answer is not true it is false

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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

lawyer [7]

(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:

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