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pickupchik [31]
3 years ago
15

Alberta Doan worked 6 hours at time-and-a-half pay and 3 ¼ hours at double-time pay. Her regular pay rate was $9.72 an hour. Wha

t was Alberta's total overtime pay for the week?
Mathematics
1 answer:
rewona [7]3 years ago
5 0

Answer:

29.22

Step-by-step explanation:

6x3/1/4+9.72

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Gemiola [76]

Answer:

t=\frac{-34.90-0}{\frac{95.66}{\sqrt{10}}}=-1.154    

The degrees of freedom are:

df = n-1= 10-1=9

The p value would be given by:

p_v =P(t_{9}

The p value is a large value and if we use a significance level of 0.05 or 0.1 we see that we can FAIL to reject the null hypothesis. And then we don't have enough evidence to conclude that her clients would save money on average by switching to the online company

Step-by-step explanation:

Information given

\bar X=-34.90 represent the mean for the difference

s=95.66 represent the sample deviation for the difference

n=10 sample size    

\mu_o =0 represent the value that we want to test  

\alpha represent the significance level

t would represent the statistic

p_v represent the p value

System of hypothesis

We want to determine if that her clients would save money on average by switching to the online company, the system of hypothesis would be:    

Null hypothesis:\mu \geq 0    

Alternative hypothesis:\mu < 0    

The statistic for this case is given by:

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

Replacing the info given we got:

t=\frac{-34.90-0}{\frac{95.66}{\sqrt{10}}}=-1.154    

The degrees of freedom are:

df = n-1= 10-1=9

The p value would be given by:

p_v =P(t_{9}

The p value is a large value and if we use a significance level of 0.05 or 0.1 we see that we can FAIL to reject the null hypothesis. And then we don't have enough evidence to conclude that her clients would save money on average by switching to the online company

4 0
4 years ago
Read 2 more answers
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