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3 years ago

One number is two less than another. Their sum is twelve. Find both numbers.

2 answers:

8 0

Hello there.

Question: <span>One number is two less than another. Their sum is twelve. Find both numbers.

Answer: It would be 5 and 7:
5 + 7 = 12
7 - 5 = 2.

Hope This Helps You!
Good Luck Studying ^-^</span>

notsponge [240]3 years ago

4 0

A = b - 2
a + b = 12
b = 7
a = 5
5 = 7 - 2
5 + 7 = 12

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Abnormalities In the 1980s, it was generally believed that congenital abnormalities affected about 5% of the nation’s chil-dren.

Ainat [17]

Answer:

Null hypothesis: $p\leq 0.05$

Alternative hypothesis: $p>0.05$

The conditions and requirements are explained on detail below.

Step-by-step explanation:

1) Data given and notation n

n=384 represent the random sample taken

X=46 represent the children with abnormalities in the sample

$\hat p=\frac{46}{384}=0.120$ estimated proportion of children with abnormalities in the sample

$p_o=0.05$ is the value that we want to test

$\alpha$ represent the significance level (no given)

z would represent the statistic (variable of interest)

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

p= proportion of children with congenital abnormalities

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion of children with congenital abnormalities exceeds 5%. :

Null hypothesis: $p\leq 0.05$

Alternative hypothesis: $p>0.05$

We assume that the proportion follows a normal distribution.

When we conduct a proportion test we need to use the z statistic, 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,higher or less) from a hypothesized value $p_o$ .

<em>Check for the assumptions that he sample must satisfy in order to apply the test</em>

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =384*0.05=19.2>10$

$n(1-p_o)=384*(1-0.05)=364.8>10$

3) Calculate the statistic

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

$z=\frac{0.120 -0.05}{\sqrt{\frac{0.05(1-0.05)}{384}}}=6.29$

4) 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 is not provided usually is $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a one side test the p value would be:

$p_v =P(z>6.29)=1.59x10^{-10}$

Based on the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of children with abnormalities exceeds 0.05 or 5% .