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Veronika [31]
3 years ago
8

juilo is paid 1.4 times his normal hourly rate for each hour he works over 30 hours in a week. Last week he worked 35hours and e

arned $436.60.Write and solve an equation to find julios's normal hourly rate. explain how you know your answer is reasonable.
Mathematics
1 answer:
Licemer1 [7]3 years ago
4 0
Let x be Juilo's normal hourly rate.

so for the 35 hr week  we have:-

30x + 5*1.4x = 436.60

37x = 436.60

x = 436.60 / 37 =  $11.80

His normal hourly rate = $11.80


So for a 30 hr week he will earn 11.80 * 30  =  $354

So  the  amount $436.60  is a reasonable figure for  a 35 hr week.
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Answer:

z=\frac{53.4-48.8}{\frac{12}{\sqrt{36}}}=2.3  

p_v =P(Z>2.3)=1-P(Z  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 48.8 at 1% of signficance.  

Step-by-step explanation:

1) Data given and notation  

\bar X=53.4 represent the sample mean

\sigma=12 represent the population standard deviation assumed

n=36 sample size  

\mu_o =48.8 represent the value that we want to test  

\alpha=0.01 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is higher than 48.8, the system of hypothesis would be:  

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Alternative hypothesis:\mu > 48  

Since we assume that know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

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z-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".  

3) Calculate the statistic  

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

z=\frac{53.4-48.8}{\frac{12}{\sqrt{36}}}=2.3  

4)P-value  

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

p_v =P(Z>2.3)=1-P(Z  

5) Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 48.8 at 1% of signficance.  

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