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pogonyaev
4 years ago
8

Estimate what is 130% of 20. Show work please! I need the right answer!

Mathematics
2 answers:
aksik [14]4 years ago
3 0
26 would 130% of 26%. 

20 would be 100% of itself leading us to X being 130%. 
That brings us to two equations of 20=100% and x=130%
20/x=100%/130%

20/x=100/130
(20/x)*x=(100/130)*x
20=0.769230769231*x
26/0.769230769231=x
26=x therefor x=26


Nata [24]4 years ago
3 0
The correct answer to your question is 26.

130% of 20 = 26.
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a) 29.4-1.96\frac{7}{\sqrt{10}}=25.06  

29.4+1.96\frac{7}{\sqrt{10}}=33.74  

So on this case the 95% confidence interval would be given by (25.06;33.74)  

b) z=\frac{29,4-25}{\frac{7}{\sqrt{10}}}=1.988  

p_v =P(z>1.988)=0.0234    

If we compare the p value and the significance level given \alpha=Step-by-step explanation:Previous conceptsA confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  
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Part a
Data given: 30 30 42 35 22 33 31 29 19 23
We can calculate the sample mean with the following formula:[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}= 29.4 the sample mean

\mu population mean (variable of interest)  

\sigma=7 represent the population standard deviation  

n=10 represent the sample size  

95% confidence interval  

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

\bar X \pm z_{\alpha/2}\frac{\sigma}{\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):  

29.4-1.96\frac{7}{\sqrt{10}}=25.06  

29.4+1.96\frac{7}{\sqrt{10}}=33.74  

So on this case the 95% confidence interval would be given by (25.06;33.74)  

Part b

What are H0 and Ha for this study?  

Null hypothesis:  \mu \leq 25  

Alternative hypothesis :\mu>25  

Compute the test statistic

The statistic for this case is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-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:  

z=\frac{29,4-25}{\frac{7}{\sqrt{10}}}=1.988  

Give the appropriate conclusion for the test

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

p_v =P(z>1.988)=0.0234  

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.    

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