Mathematics: Help me solve for x please

Help me solve for x please - id14009203, kiyahlove7537

Ответ:

priushunter9463

03.08.2022

Mathematics

a) $1.30-2.306\frac{0.2}{\sqrt{9}}=1.146$

$1.30+2.306\frac{0.2}{\sqrt{9}}=1.454$

So on this case the 95% confidence interval would be given by (1.146;1.454)

b) $n=(\frac{2.326(0.2)}{0.065})^2 =51.22 \approx 52$

So the answer for this case would be n=52 rounded up to the nearest integer

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".

$\bar X=1.3$ represent the sample mean

$\mu$ population mean (variable of interest)

s=0.2 represent the sample standard deviation

n=9 represent the sample size

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)

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

df=n-1=9-1=8

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(0.025,8)".And we see that $t_{\alpha/2}=2.306$

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

$1.30-2.306\frac{0.2}{\sqrt{9}}=1.146$

$1.30+2.306\frac{0.2}{\sqrt{9}}=1.454$

So on this case the 95% confidence interval would be given by (1.146;1.454) Part b

The margin of error is given by this formula:

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

And on this case we have that ME =0.05*1.3= 0.065 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 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.01;0;1)", and we got $z_{\alpha/2}=2.326$ , replacing into formula (b) we got:

$n=(\frac{2.326(0.2)}{0.065})^2 =51.22 \approx 52$

So the answer for this case would be n=52 rounded up to the nearest integer

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