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sergey [27]
3 years ago
8

Terrell needs the total weight of his backpack, water jug, and sleeping bag to be less than 11 kilograms. His backpack weighs 9,

580 grams, his water jug is 1.25 kilograms, and his sleeping bag is 25,000 milligrams. Dominic says that the total weight is 35.83 kg. He shows Terrell his math.
Backpack: 9,580÷ 1,000= 9.58 kg total 9.58
water jug 1.25 kg 1.25
Sleeping bag 25,000÷1,000= 25 kg +25.00
______
35.83
a. What mistake did Dominic make?
Mathematics
1 answer:
Ket [755]3 years ago
5 0
He did not divide his milligrams by the proper factor. He treated it as if it were only kilograms. He should have divided by a million. 
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Answer:

t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37  

p_v =P(t_{4}>2.37)=0.038  

If we compare the p value and a significance level assumed \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.  

Step-by-step explanation:

Data given and notation

\bar X=250000 represent the sample mean  

s=37500 represent the standard deviation for the sample

n=5 sample size  

\mu_o =210250 represent the value that we want to test  

\alpha=0.1 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 to be tested  

We need to conduct a hypothesis in order to determine if the mean is higher than 210250, the system of hypothesis would be:  

Null hypothesis:\mu \leq 210250  

Alternative hypothesis:\mu > 210250  

Compute the test statistic  

We don't know the population deviation, so for this case 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{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37  

Now we need to find the degrees of freedom for the t distirbution given by:

df=n-1=5-1=4

Conclusion

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

p_v =P(t_{4}>2.37)=0.038  

If we compare the p value and a significance level assumed \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.  

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