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netineya [11]
3 years ago
15

PLEEASE COME TO MY CLASS AND CRASH IT WITH DN JOKES OR MUSIC

Mathematics
2 answers:
Grace [21]3 years ago
3 0
Ok bet I got you say less
zloy xaker [14]3 years ago
3 0

Sheeeeeeeshhhhhh say less

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I need help with this there is part a and b
Alex17521 [72]

Answer:

the answer should be 3.914 or 4

Step-by-step explanation:

hope it helps!

I am guessing that Part A: 3.914

and                          Part B: 4

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Please answer! Will give brainlest!
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11) the product of 77 and h

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Please help me if you know the answer and can you explain it to me
Slav-nsk [51]

Answer:

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Step-by-step explanation:

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3 years ago
To evaluate the effect of a treatment, a sample is obtained from a population, with a mean of u = 30, and the treatment is admin
Murrr4er [49]

Answer:

a) t=\frac{31.3-30}{\frac{3}{\sqrt{16}}}=1.733  

p_v =2*P(t_{15}>1.733)=0.104  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the we don't have a significant effect for the new treatment at 5% of significance.  

b) t=\frac{31.3-30}{\frac{3}{\sqrt{36}}}=2.6  

p_v =2*P(t_{15}>2.6)=0.0201  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the we have a significant effect for the new treatment at 5% of significance.  

c) When we increase the sample size we increse the probability of rejection of the null hypothesis since the z score tend to increase when the sample size increase.

Step-by-step explanation:

Data given and notation  

Part a: If the sample consists of n=16 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha = 0.05.

\bar X=31.3 represent the sample mean  

s=3 represent the sample standard deviation  

n=16 sample size  

\mu_o =30 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 30, the system of hypothesis are :  

Null hypothesis:\mu = 30  

Alternative hypothesis:\mu \neq 30  

Since we don't know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{31.3-30}{\frac{3}{\sqrt{16}}}=1.733  

P-value  

First w eneed to find the degrees of freedom given by:

df=n-1=16-1 =15

Since is a two-sided test the p value would given by:  

p_v =2*P(t_{15}>1.733)=0.104  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the we don't have a significant effect for the new treatment at 5% of significance.  

Part b: If the sample consists of n=36 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha = 0.05.

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{31.3-30}{\frac{3}{\sqrt{36}}}=2.6  

P-value  

First w eneed to find the degrees of freedom given by:

df=n-1=16-1 =15

Since is a two-sided test the p value would given by:  

p_v =2*P(t_{15}>2.6)=0.0201  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the we have a significant effect for the new treatment at 5% of significance.  

Part c; Comparing your answer for parts a and b, how does the size of the sample influence the outcome of a hypothesis test

When we increase the sample size we increse the probability of rejection of the null hypothesis since the z score tend to increase when the sample size increase.

5 0
3 years ago
Examine the worked problem and solve the equation.
ANTONII [103]

Answer:

The answer is 7

Step-by-step explanation:

I don't know how to explain it really but just divide 4/3 by 28/3 and you should get the answer of x

5 0
3 years ago
Read 2 more answers
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