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KengaRu [80]
3 years ago
7

How do you use proportions to solve percent problems​

Mathematics
1 answer:
exis [7]3 years ago
7 0

Answer:

Step-by-step explanation:

EXAMPLE #1:

What number is 75% of 4?   (or   Find 75% of 4.)

The PERCENT always goes over 100.

   (It's a part of the whole 100%.)

4 appears with the word of:

   It's the WHOLE and goes on the bottom.

A proportion showing one fraction  with PART as the numerator and 4 as the denominator equal to another fraction with 75 as the numerator and 100 as the denominator.

We're trying to find the missing PART (on the top).

In a proportion the cross-products are equal:   So 4 times 75 is equal to 100 times the PART.

The missing PART equals 4 times 75 divided by 100.

(Multiply the two opposite corners with numbers; then divide by the other number.)

4 times 75 = 100 times the part

300 = 100 times the part

300/100 = 100/100 times the part

3 =   the part

A proportion showing the denominator,  4, times the diagonally opposite 75; divided by  100.

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sergejj [24]

Answer:

a) Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

b) t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958      

c) p_v =P(t_{34}>2.958)=0.0028    

d) If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

The sample mean is given by:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And the sample deviation is:

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=17 represent the sample mean    

s=4 represent the sample standard deviation

n=35 sample size  

\mu_o =15 represent the value that we want to test  

\alpha=0.01 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

Part a State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less or equal than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

Since we don't  know the population deviation, 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".  

Part b Calculate the statistic  

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

t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958  

Part c P-value  

The degrees of freedom are given by:

df = n-1= 35-1=34

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

p_v =P(t_{34}>2.958)=0.0028  

Part d Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 1% of signficance.  

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