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Komok [63]
4 years ago
12

drew's weekly allowance is $8.00.His friend Jann' s weekly allowance is $10 Drew spends $3 a week and Jan spends $4 a week. Writ

e two expressions to show how much money each person has at the end of the week?
Mathematics
1 answer:
lidiya [134]4 years ago
3 0
8-3=$5
10-4=$6
That's how much each person has.
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x>=0

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4 years ago
In a recent Super Bowl, a TV network predicted that 50 % of the audience would express an interest in seeing one of its forthcom
coldgirl [10]

Answer:

z= -0.968

We can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that  they would watch one of the television shows not differs from 0.5 or 50% .  

Step-by-step explanation:

1) Data given and notation n  

n=106 represent the random sample taken

X=48 represent the people who says that  they would watch one of the television shows.

\hat p=\frac{48}{106}=0.453 estimated proportion of people who says that  they would watch one of the television shows.

p_o=0.5 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 50% of people who says that  they would watch one of the television shows.:  

Null hypothesis:p=0.5  

Alternative hypothesis:p \neq 0.5  

When we conduct a proportion test we need to use the z statisitc, 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{0.453 -0.5}{\sqrt{\frac{0.5(1-0.5)}{106}}}=-0.968  

4) Statistical decision  

P value method or p value approach . "This method consists on 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 is not provided, but we can assume \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z  

So based on the p value obtained and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that  they would watch one of the television shows not differs from 0.5 or 50% .  

6 0
4 years ago
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