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AlexFokin [52]
3 years ago
12

Can someone help me with this one

Mathematics
1 answer:
Nostrana [21]3 years ago
6 0
It’s name would be a monomial because it only has 1 term. (Terms are departed by addition or subtraction sign) The degree would be the highest exponent in the polynomial, in this case we only have 1 term so the degree of that term is 2
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Solve for x. 7^8=7^8×7^x
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Read 2 more answers
The mean time required to repair breakdowns of a certain copying machine is 93 minutes. The company which manufactures the machi
Sonja [21]

Answer:

t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349    

p_v =P(t_{(72)}  

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 reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=88.8 represent the sample mean

s=26.6 represent the sample standard deviation for the sample  

n=73 sample size  

\mu_o =93 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 true mean i lower than 93 min, the system of hypothesis would be:  

Null hypothesis:\mu \geq 93  

Alternative hypothesis:\mu < 93  

If we analyze the size for the sample is > 30 but 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{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349    

P-value

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

df=n-1=73-1=72  

Since is a one side test the p value would be:  

p_v =P(t_{(72)}  

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 reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.  

5 0
3 years ago
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