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3 years ago

Average of 6 numbers is 4, if the average of 2 numbers is 2 what is the average of other 4

Mathematics

1 answer:

RideAnS [48]3 years ago

4 0

Answer:

Step-by-step explanation:

Let's just pretend the 6 numbers are: a,b,c,d,e,f.

Then (a+b+c+d+e+f)/6 =4 (given by first statement.)

Let's pretend the average of 2 numbers is 2 means they are talking about the a and b.

So (a+b)/2=2.

Now we want to find (c+d+e+f)/4 as requested by their question.

So first step multiply both sides of our first equation by 6 giving us:

a+b+c+d+e+f=24

So the second step multiply both sides of our second equation by 2 giving us:

a+b=4.

Now if a+b+c+d+e+f=24 and a+b=4, then c+d+e+f=20 since 20+4=24.

So the average of the four numbers c,d,e, and f is 20/4=5.

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bagirrra123 [75]

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$z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363$

$p_v =P(z>3.363)=0.00039$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

Step-by-step explanation:

Data given and notation

n=1655 represent the random sample taken

$\hat p=0.021$ estimated proportion of interest

$p_o=0.012$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportions is higher than 0.012.:

Null hypothesis: $p \leq 0.012$

Alternative hypothesis: $p > 0.012$

When we conduct a proportion test we need to use the z statisitic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>3.363)=0.00039$

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