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3 years ago

Kareem wrote the values shown below.

Mathematics

1 answer:

grandymaker [24]3 years ago

7 0

Answer:

Option D is correct.

$\lfloor 4.8 \rfloor$ and $\lceil 3.2 \rceil$

Step-by-step explanation:

Floor function states that the greatest integer that is less than or equal to x.

It is represented by: $\lfloor x \rfloor$

Ceiling function states that the least integer that is greater than or equal to x.

It is represented by: $\lceil x \rceil$

As per the statement:-

Kareem wrote the values show below:

$\lceil 3.2 \rceil$ , $\lfloor 2.7 \rfloor$ , $\lceil 2.9 \rceil$ and $\lfloor 4.8 \rfloor$ .

by definition:

$\lceil 3.2 \rceil$ = 4

$\lfloor 2.7 \rfloor$ = 2

$\lceil 2.9 \rceil$ =3

$\lfloor 4.8 \rfloor$ = 4

From the given options only pair which is equivalent is:

$\lfloor 4.8 \rfloor$ and $\lceil 3.2 \rceil$

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put(3) in(2)

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$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

Null hypothesis: $p\geq 0.8$

Alternative hypothesis: $p < 0.8$

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

$p_v =P(Z$

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

$p_v =P(Z>2.124)=0.013$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

n=110 represent the random sample taken

X=97 represent the students who completed the modules before class

$\hat p=\frac{97}{110}=0.882$ estimated proportion of students who completed the modules before class

$p_o=0.8$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest) 2) Concepts and formulas to use We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.8 or 80%: Null hypothesis:[tex]p\geq 0.8$

Alternative hypothesis: $p < 0.8$

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 from a hypothesized value $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

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 assumed $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(Z$

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

$p_v =P(Z>2.124)=0.013$

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