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iren2701 [21]
3 years ago
12

2. How many ounces of Ceylon tea worth $1.50 per ounce

Mathematics
1 answer:
gregori [183]3 years ago
4 0

Answer: How many ounces of ceylon tea worth 1.50 per ounce and how many ounces of formosa tea worth 2.00 per ounce must be mixed to obtain a mixture of 8 ounces that is worth 1.85 per ounce

Equation:

1.5c + 2f = 1.85*8

c + f = 8

Modify:

1.5c + 2f = 1.85*8

2c + 2f = 16

Subtract and solve for "c":

0.5c = 1.2

c = 2.4 ounces (amt. of ceylon tea needed)

Solve for "f":: c + f = 8, so f = 5.2 ounces

Step-by-step explanation:

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Answer:

t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744    

p_v =P(t_{(47)}>2.744)=0.0043  

If we compare the p value and the significance level given \alpha=0.02 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=2.1 represent the mean

s=1.01 represent the sample standard deviation

n=48 sample size  

\mu_o =1.7 represent the value that we want to test

\alpha=0.02 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 higher than 1.7, the system of hypothesis would be:  

Null hypothesis:\mu \leq 1.7  

Alternative hypothesis:\mu > 1.7  

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{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744    

P-value

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

df=n-1=48-1=47  

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

p_v =P(t_{(47)}>2.744)=0.0043  

Conclusion  

If we compare the p value and the significance level given \alpha=0.02 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.  

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