In how many ways can you select five people from a group of nine if the order of selection is important?

1 answer:

4 0

If the order of the selection is important we will apply the Arrangements formula $A_{n} ^{k}= \frac{n!}{(n-k)!}$ .
In our case we will have $A_{9}^{5}= \frac{9!}{4!} =5*6*7*8*9=15120.$

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You have measured the systolic blood pressure of a random sample of 25 employees of a company located near you. A 95% confidence

Alla [95]

Answer:

$122 \leq \mu \leq 138$

The 95% confidence interval would be given by (122;138)

And the correct interpretation for this case is:

c. If the procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure.

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 % where by 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".

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

$df=n-1=25-1=24$

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(0.025,24)".And we see that $t_{\alpha/2}=2.06$