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PSYCHO15rus [73]
3 years ago
7

Joni sold 376 tickets for the school play. Student tickets cost $4 and adult tickets cost $7. Joni's sales totaled $2080.

Mathematics
1 answer:
nikklg [1K]3 years ago
5 0

Answer:

Number of students tickets sold= 184

Number of adults tickets sold= 192

Step-by-step explanation:

I assume we need to calculate the number of student and adult tickets sold.

<u>First, we establish the system of equations:</u>

x= number of students tickets sold

y= number of adults tickets sold

4*x + 7*y= 2,080

x + y= 376

<u>Now, we isolate x in one formula and substitute it in the other:</u>

x= 376 - y

4*(376 - y) + 7y = 2,080

1,504 - 4y + 7y = 2,080

3y = 576

y= 192

x= 376 - 192

x= 184

Number of students tickets sold= 184

Number of adults tickets sold= 192

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According to one investigator’s model, the data are like 400 draws made at random from a large box. The null hypothesis says tha
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Answer:

z=\frac{52.75-50}{\frac{25}{\sqrt{400}}}=2.2    

p_v =2*P(Z>2.2)=0.0278  

If we compare the p value and the significance level assumed \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the mean is significantly diffrent fro 50 at 1% of signficance.  

Step-by-step explanation:

1) Data given and notation  

\bar X=52.75 represent the mean height for the sample  

s=25 represent the sample standard deviation

n=400 sample size  

\mu_o =50 represent the value that we want to test

\alpha 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 equal to 50 or not, the system of hypothesis would be:  

Null hypothesis:\mu = 50  

Alternative hypothesis:\mu \neq 50  

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 but we can assume it as z distribution, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{s}{\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".  

Calculate the statistic

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

z=\frac{52.75-50}{\frac{25}{\sqrt{400}}}=2.2    

P-value

Since is a two sided test the p value would be:  

p_v =2*P(Z>2.2)=0.0278  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the mean is significantly diffrent fro 50 at 1% of signficance.  

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