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jekas [21]
3 years ago
7

I need help on this question!

Mathematics
2 answers:
zlopas [31]3 years ago
6 0
The answer I am not sure but will try
Ans: 110
bija089 [108]3 years ago
6 0

Answer:

110 cookies in 180 minutes

Step-by-step explanation:

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The next term of the sequence is 48.<br> (2, 4, 8, 16, 32,...)
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In a sample of 170 students at an Australian university that introduced the use of plagiarism-detection software in a number of
Kisachek [45]

Answer:

p_v =P(Z>4.146)=0.0000169  

Based on the p value obtained and using the significance level given \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 students who NOT belief that such software unfairly targets students is higher than 0.5 or 50% .  

Step-by-step explanation:

1) Data given and notation  

n=170 represent the random sample taken  

X=58 represent the student's who belief that such software unfairly targets students

\hat px=\frac{58}{170}=0.341 estimated proportion of students who belief that such software unfairly targets students

\hat p=\frac{112}{170}=0.659 estimated proportion of students who NOT belief that such software unfairly targets students

p_o=0.50 is the value that we want to test  

\alpha=0.05 represent the significance level (no given)  

z would represent the statistic (variable of interest)  

p_v represent the p value (variable of interest)  

p= proportion of student's who belief that such software unfairly targets students

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that that a majority of students at the university do not share this belief. :  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p>0.5

We assume that the proportion follows a normal distribution.  

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,higher or less) from a hypothesized value p_o.  

<em>Check for the assumptions that he sample must satisfy in order to apply the test </em>

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

np_o =170*0.5=85>10

n(1-p_o)=170*(1-0.5)=85>10

3) Calculate the statistic  

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

z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{170}}}=4.146  

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

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

p_v =P(Z>4.146)=0.0000169  

Based on the p value obtained and using the significance level given \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 students who NOT belief that such software unfairly targets students is higher than 0.5 or 50% .  

8 0
2 years ago
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