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atroni [7]
2 years ago
14

There are 7 times as many males as females on the maths course at university. What fraction of the course are male? Give your an

swer in its simplest form.
Mathematics
1 answer:
podryga [215]2 years ago
3 0

Answer:

The fraction of the course 7/8 is male on the maths course at university.

Step-by-step explanation:

There are 7 times as many males as females on the maths course at University. Let's suppose we have 10 females in the course.If we have 10 females, the males should be 7 times; it means with 10 females males should be 70 males.

Now, we have the number of males (M) - 70 and females (F) - 10. We can calculate the fraction of the male (%M) in the course.

%M = M/(M+F)

%M = 70 / (70 +10)

% M = 70/80

%M = 7/8.

Finally, we can conclude that the fraction of the course 7/8 are male on the maths course at university. 

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Null hypothesis:p \leq 0.69  

Alternative hypothesis:p > 0.69  

Step-by-step explanation:

1) Data given and notation

n=1500 represent the random sample taken

X represent the number of graduates that had student loan debt in 2014

\hat p=0.71 estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

p_o=0.69 is the value that we want to test

\alpha represent the significance level  

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that if there was a significant increase in the proportion of student loan debt for public and nonprofit colleges in 2014 respect to the value of 2013.:  

Null hypothesis:p \leq 0.69  

Alternative hypothesis:p > 0.69  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.71 -0.69}{\sqrt{\frac{0.69(1-0.69)}{1500}}}=1.675  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is \alpha=0.05. The next step would be calculate the p value for this test.  

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

p_v =P(Z>1.675)=0.047  

If we compare the p value obtained and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults that said that it is morally wrong to not report all income on tax returns  is not significantly higher than 0.69.  

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