Find the solutions please help me

77julia77 [94]

You turn 4/5 to 8/10 to get equal denominators and add the numerators together to get 16/10, or 1 6/10 which is reduced to 1 3/5.

8 0

3 years ago

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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

4 years ago

A gambler claims that horses in a horse race that start closer to the rail have an advantage over horses that start further from

k0ka [10]

Answer:

Hence, the p-value ( 0.699986) > 0.05, FAIL TO REJECT THE NULL HYPOTHESIS

there is no significant evidence that starting position is a factor in horse races

Step-by-step explanation:

Given the data in the question;

Starting position: 1 2 3 4 5 6

# of Winners: 25 22 18 19 21 15

Mean x" (E) = (25+22+18+19+21+15)/6 = 120/6 = 20

Observed/Expected value table;

O E ((O-E)²) / E

25 20 1.25

22 20 0.2

18 20 0.2

19 20 0.05

21 20 0.05

15 20 1.25

∑ 3

so using chi square = ∑(((O-E)²) / E) = 3

DF = n-1 = 6-1 = 5

then, the critical chi square value is;

significance level = 0.05

chi square( critical ) = 11.07049769

from table;

p - value = 0.699986

Hence, the p-value ( 0.699986) > 0.05, FAIL TO REJECT THE NULL HYPOTHESIS

there is no significant evidence that starting position is a factor in horse races

8 0

3 years ago

Which expression is equivalent to n2 26n 88 for all values of n? (n 8)(n 11) (n 4)(n 22) (n 4)(n 24) (n 8)(n 18)

Sunny_sXe [5.5K]

The factor of the given polynomial n² + 26n + 88 is (n + 4) and (n + 22). Then the correct option is B.

<h3>What is a factorization?</h3>

It is the method to separate the polynomial into parts and the parts will be in multiplication. And the value of the polynomial at this point will be zero.

The polynomial is given below.

n² + 26n + 88

Then the factor of the given polynomial will be

n² + 22n + 4n + 88

n(n + 22) 4(n + 22)

(n + 4)(n + 22)

More about the factorization link is given below.

brainly.com/question/6810544

8 0

2 years ago

I WILL REWARD BRAINLIEST

aksik [14]

Answer:

Answer: What is the total number of students who brought water bottles to lunch?

A statistical question is a question with data that will vary.

4 0

3 years ago

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