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chubhunter [2.5K]
3 years ago
8

Plz help in my last question​

Mathematics
2 answers:
Neko [114]3 years ago
5 0

Answer:

Where is the question??

Greeley [361]3 years ago
3 0

Answer:

where is the question, lady?

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If the critical z-value for a hypothesis test... If the critical z-value for a hypothesis test equals 2.45, what value of the te
schepotkina [342]

Answer: z score = 0.00714

Step-by-step explanation: the value of test statistics is gotten using the standard normal distribution table.

Z= 2.45 has area to the left (z<2.45) and area to the right (z>2.45).

Level of significance α is the probability of committing a type 1 error. The area under the distribution is known as the rejection region and it is the area towards the right of the distribution.

The table I'm using is towards the left of the distribution.

But z>2.45 + z<2.45 = 1

z> 2.45 = 1 - z<2.45

But z < 2.45 = 0.99286

z > 2.45 = 1 - 0.99286

z >2.45 = 0.00714

Hence the test statistics that would produce the least type 1 error is 0.00714

8 0
3 years ago
Which ratio is equivalent to the ratio 33:36
xenn [34]

Answer:

Step-by-step explanation:

33/3 = 11

36/3 =13

11 :13

5 0
3 years ago
Question b with explanation?​
Step2247 [10]
Hello!

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8 0
2 years ago
Consider the computer output below. Fill in the missing information. Round your answers to two decimal places (e.g. 98.76). Test
slamgirl [31]

Answer:

SE_{Mean}=\frac{s}{\sqrt{n}}=\frac{4.77}{\sqrt{19}}=1.094

t=\frac{98.77-100}{\frac{4.77}{\sqrt{19}}}=-1.124      

The 95% confidence interval would be given by (96.625;100.915)  

a) df=n-1= 19-1= 18

b) p_v =2*P(t_{18}      

If we compare the p value and a significance level for example \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.

c) The only thing that changes is the p value and would be given by:

p_v =P(t_{18}>-1.124)=0.862      

But again since the p value is higher than the significance level we fail to reject the null hypothesis.

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

\bar X=98.77 represent the sample mean    

s=4.77 represent the sample standard deviation  

n=19 represent the sample selected  

\alpha significance level    

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if we have significant difference on the mean of 100, the system of hypothesis would be:    

Null hypothesis:\mu = 100    

Alternative hypothesis:\mu \neq 100    

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we can calculate the Standard error for the mean like this:

SE_{Mean}=\frac{s}{\sqrt{n}}=\frac{4.77}{\sqrt{19}}=1.094

The confidence interval for the mean is given by the following formula:  

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96  

Now we have everything in order to replace into formula (1):  

98.77-1.96\frac{4.77}{\sqrt{19}}=96.625  

98.77+1.96\frac{4.77}{\sqrt{19}}=100.915  

So on this case the 95% confidence interval would be given by (96.625;100.915)  

Part a

The degree of freedom are given by:

df=n-1= 19-1= 18

Part b

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t 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".    

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

t=\frac{98.77-100}{\frac{4.77}{\sqrt{19}}}=-1.124      

Then since is a two sided test the p value would be:    

p_v =2*P(t_{18}      

If we compare the p value and a significance level for example \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.

Part c

If the system of hypothesis on this case are:

Null hypothesis:\mu = 100    

Alternative hypothesis:\mu > 100  

The only thing that changes is the p value and would be given by:

p_v =P(t_{18}>-1.124)=0.862      

But again since the p value is higher than the significance level we fail to reject the null hypothesis.

7 0
3 years ago
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