Mathematics, 23.12.2019 22:31 braydentillery1221

Suppose that the researchers wanted to estimate the mean reaction time to within 6 msec with 95% confidence. using the sample standard deviation from the study described as a preliminary estimate of the standard deviation of reaction times, compute the required sample size. (round your answer up to the nearest whole number.)

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Answer from: cheerkennedym2594

a) The 95% confidence interval would be given by (509.592;550.308)

b) n=523 rounded up to the nearest integer

Step-by-step explanation:

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".

$\bar X=530$ represent the sample mean for the sample

$\mu$ population mean

s=70 represent the sample standard deviation

n=48 represent the sample size (variable of interest)

Confidence =95% or 0.995

Part a

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

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

The degrees of freedom are df=n-1=48-1=47

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 table to find the critical value. The excel command would be: "=-T.INV(0.025,47)".And we see that $z_{\alpha/2}=2.01$

Now we have everything in order to replace into formula (1):

$530-2.01\frac{70}{\sqrt{48}}=509.692$

$530+2.01\frac{70}{\sqrt{48}}=550.308$

So on this case the 95% confidence interval would be given by (509.592;550.308)

Part b

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Assuming that $\hat \sigma =s$

And on this case we have that ME =6msec, and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval is provided, $z_{\alpha/2}=1.96$ , replacing into formula (b) we got:

$n=(\frac{1.96(70)}{6})^2 =522.88 \approx 523$

So the answer for this case would be n=523 rounded up to the nearest integer

Answer from: Quest

f(2x−x2)=−3x2+6x−1

step-by-step explanation:

set up the composite function and evaluate.

f(2x−x2)=−3x2+6x−1

Answer from: Quest

Gggģcawrygr3ezt9d hu and the other countries have been working to make it easier to box up their homes in an area where the government has not yet decided which countries will have the option of the government

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