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oksian1 [2.3K]
3 years ago
8

I need to kno the answer

Mathematics
1 answer:
ohaa [14]3 years ago
7 0
Sorry I can see it but thank for your question.
You might be interested in
Find the left critical value for 95% confidence interval for σ with n = 41. 26.509 24.433 55.758 59.342
Makovka662 [10]

Answer: 59.342

Step-by-step explanation:

The chi-square critical values are used to find the confidence interval for σ.

Left critical value = \chi^2_{\alpha/2, n-1} [i.e. chi-square value from chi-square table corresponding to degree of freedom n-1 and significance level of \alpha/2]

To find : left critical value for 95% confidence interval for σ with n = 41.

Significance level : \alpha=1-0.95=0.05

degree of freedom = 41-1=40

Now, the  left critical value for 95% confidence interval for σ with n = 41 is the chi-square value corresponding  to degree of freedom n-1 and \alpha/2=0.025

=59.342  [from chi-square table ]

7 0
3 years ago
4x + y = 46.75 help!!<br> A 8.50<br> B 8.75<br> C 9.50<br> D12.75
WARRIOR [948]

Answer:

d is the answer

Step-by-step explanation:

3 0
3 years ago
A counselor at a community college claims that the mean GPA for students who have transferred to State University is more than 3
Lena [83]

Answer:

t=\frac{3.25-3.1}{\frac{0.3}{\sqrt{36}}}=3  

If we compare the p value and the significance level for example \alpha=1-0.99=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.

And the best conclusion would be:

a) Yes. The counselor's claim is valid

Because we reject the null hypothesis in favor to the alternative hypothesis.

Step-by-step explanation:

Data given and notation  

\bar X=3.25 represent the sample mean

s=0.3 represent the sample standard deviation

n=36 sample size  

\mu_o =3.1 represent the value that we want to test  

\alpha 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 more than 3.1, the system of hypothesis would be:  

Null hypothesis:\mu \leq 3.1  

Alternative hypothesis:\mu > 3.1  

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".  

Calculate the statistic  

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

t=\frac{3.25-3.1}{\frac{0.3}{\sqrt{36}}}=3  

P-value  

The first step is calculate the degrees of freedom, on this case:  

df=n-1=36-1=35  

Since is a one right tailed test the p value would be:  

p_v =P(t_{(35)}>3)=0.00247  

Conclusion  

If we compare the p value and the significance level for example \alpha=1-0.99=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.

And the best conclusion would be:

a) Yes. The counselor's claim is valid

Because we reject the null hypothesis in favor to the alternative hypothesis.

5 0
3 years ago
There were 10 bunches of roses and tulips. Each bunch had 5 of the same flowers. There were 2 more bunches of tulips than roses.
Leviafan [203]

Answer:

I think the answer is 12 (sorry if i misunderstood the question)

Step-by-step explanation:

Because if there are 10 things of roses and tulips and then 2 more tulips are added in the question in total thats 12.... 10+2=12

6 0
3 years ago
a recipe for chicken and rice calls for 3 and 1/2 pounds of chicken. Lisa wants to adjust the recipe so that it yields 1 and 1/2
elena-14-01-66 [18.8K]
I'm going to put it into decimals because it is easier. 

3.5*1.5=5.25 or 5 \frac{1}{4}

Hope that helps
3 0
2 years ago
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