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Mazyrski [523]
3 years ago
11

.Sally is comparing 2 types of house insurance. Both companies provide the same service; however,

Mathematics
1 answer:
Nostrana [21]3 years ago
6 0

Answer:

car help

Step-by-step explanation:

In order to determine which type of house insurance would be considered as cheapest first it is necessary to determine either monthly or either weekly

As we know that

Generally in the month there is a 4 weeks

So For Car Help, the monthly would be

= $8.60 × 4

= $34.40

And for Car safety, the quoted price is $36.10 per week

So according to the above calculation the car help would be choose as the cheapest

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Question 20 (5 points)
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The national mean annual salary for a school administrator is $90,00 a year (The Cincinnati Enquirer, April 7, 2012). A school o
timurjin [86]

Answer:

a) Null hypothesis:\mu = 90000  

Alternative hypothesis:\mu \neq 90000  

b) p_v =2*P(t_{(24)}  

c) If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean for the salary differs from 9000 at 5% of significance.

Step-by-step explanation:

1) Data given and notation  

77600 ,76000 ,90700 ,97200 ,90700 ,101800 ,78700 ,81300 ,84200 ,97600 ,

77500 ,75700 ,89400 ,84300 ,78700 ,84600 ,87700 ,103400 ,83800 ,101300

94700 ,69200 ,95400 ,61500 ,68800

We can calculate the sample mean and deviation with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

The values obtained are:

\bar X=85272 represent the mean annual salary for the sample  

s=11039.23 represent the sample standard deviation for the sample  

n=25 sample size  

\mu_o =90000 represent the value that we want to test

\alpha=0.05 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)  

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean salary differs from 90000, the system of hypothesis would be:  

Null hypothesis:\mu = 90000  

Alternative hypothesis:\mu \neq 90000  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-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".  

Part b: Calculate the statistic

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

t=\frac{85272-90000}{\frac{11039.23}{\sqrt{25}}}=-2.141    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=25-1=24  

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

p_v =2*P(t_{(24)}  

Part c: Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean for the salary differs from 9000 at 5% of significance.

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The next model of a sports car will cost 4.7% more than the current model. The current model costs $48,000. How much will the pr
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Answer:

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price - $50256

Step-by-step explanation:

multiply the percent (in decimal form) by the original cost

48000(0.47)=2256

the increase is $2256

to find the price of it, add the two prices together.

48000+2256=50256

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