1.3-2.1g when g is 1.9. Can you please help? I keep getting 2.69 but the paper shows the answer to be 5.29

Imagine that you have plotted many data points on an xy-plane. Your points seem to align into a clear best-fit line. Do you thin

sattari [20]

It depends really. If you stay close to the present, then predicting future results isn't too bad. The further you go out, the more unpredictable things get. This is because the points may deviate from the line of best fit (aka regression line) as time wears on. Of course, it also depends on what kind of data we're working with. Some pairs of variables are naturally going to correlate very strongly together. An example would be temperature versus ice cream sales.

5 0

4 years ago

Scores on the SAT Mathematics test are believed to be normally distributed. The scores of a simple random sample of five student

AysviL [449]

Answer:

The mean calculated for this case is $\bar X=584$

And the 95% confidence interval is given by:

$584-2.776\frac{86.776}{\sqrt{5}}=476.271$

$584+2.776\frac{86.776}{\sqrt{5}}=691.729$

So on this case the 95% confidence interval would be given by (476.271;691.729)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=584$

The sample deviation calculated $s=86.776$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=5-1=4$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that $t_{\alpha/2}=2.776$