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Brums [2.3K]
2 years ago
10

Lara borrowed $4500 for some home repairs. She will be paying 6% in simple interest over the next three years. What is the total

amount she will be paying on the loan
Mathematics
2 answers:
Elodia [21]2 years ago
6 0

Step-by-step explanation:

madam [21]2 years ago
3 0

Answer:

5310

Step-by-step explanation:

6% of 4500=270

270x3years=810

4500+810=5310

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The amount spent on food and drink by 20 customers is shown below What is the modal amount spent and the median?
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2 years ago
In 2009, the Southeastern Conference (SEC) commissioner set a goal to have greater than 65% of athletes that are entering freshm
dangina [55]

Answer:

z=\frac{0.7 -0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}=1.048  

p_v =P(z>1.048)=0.147  

So the p value obtained was a very high value and using the significance level given \alpha=0.1 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of people who graduate in 6 years is not significantly higher than 0.65 .  

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

X=70 represent the number of people who graduate in 6 years

\hat p=\frac{70}{100}=0.7 estimated proportion of people who graduate in 6 years

p_o=0.65 is the value that we want to test

\alpha=0.1 represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that true proportion is higher than 0.65:  

Null hypothesis:p \leq 0.65  

Alternative hypothesis:p >0.65  

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.

Calculate the statistic  

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

z=\frac{0.7 -0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}=1.048  

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 \alpha=0.1. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =P(z>1.048)=0.147  

So the p value obtained was a very high value and using the significance level given \alpha=0.1 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of people who graduate in 6 years is not significantly higher than 0.65 .  

7 0
3 years ago
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