HELP PLEASE What is M CEA?

Helppppppppppppp plssssssssssssssssssssssss

Goshia [24]

I would but I’m stuck on this problem too

3 0

2 years ago

One sample has Aa Aa One sample has n 10 scores and a variance of s2 20, and a second sample has n 15 scores and a variance of s

siniylev [52]

Answer:

option (a) It will be closer to 30 than to 20

Step-by-step explanation:

Data provided in the question:

For sample 1:

n₁ = 10

variance, s₁² = 20

For sample 2:

n₂ = 15

variance, s₂² = 30

Now,

The pooled variance is calculated using the formula,

$S^{2}_{p} = \frac{(n_{1}-1)\times s^{2}_{1} +(n_{2}-1)\times s^{2}_{2}}{n_{1}+n_{2}-2}$

on substituting the given respective values, we get

$S^{2}_{p} = \frac{(10-1)\times 20 +(15-1)\times 30}{10+15-2}$

or

$S^{2}_{p}$ = 26.0869

Hence,

the pooled variance will be closer to 30 than to 20

Therefore,

The correct answer is option (a) It will be closer to 30 than to 20

4 0

3 years ago

Simplify x^2+3x-10/x^2-25

Natasha_Volkova [10]

Answer:

3x -10 / -25

Step-by-step explanation:

cancel off the x²

6 0

3 years ago

We are conducting a hypothesis test to determine if fewer than 80% of ST 311 Hybrid students complete all modules before class.

Juli2301 [7.4K]

Answer:

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

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:

$p_v =P(Z$

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.

5 0

3 years ago

Four different sets of objects contain 2, 5, 6, and 7 objects, respectively. How many unique combinations can be formed by picki

mario62 [17]

The correct answer for the question that is being presented above is this one: "A. 960."Four different sets of objects contain 2, 5, 6, and 7 objects, respectively.

= 2 * 5 * 6 * 7
= 960

Here are the following choices:
A. 960
B. 141 
C. 23 
D. 529

3 0

3 years ago

Read 2 more answers