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adoni [48]
3 years ago
6

om is the deli manager at a grocery store. He needs to schedule employees to staff the deli department at least 260 person-hours

per week. Tom has one part-time employee who works 20 hours per week. Each full-time employee works 40 hours per week.
Mathematics
1 answer:
qwelly [4]3 years ago
8 0
20 + 40x = 260
40x = 260 - 20 
40x = 240
x = 240/40
x = 6 fulltime employees needed
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Answer:

z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{75}}}=2.43  

Now we can calculate the p value with the following probability:

p_v =P(z>2.43)=0.0075 \approx 0.008  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion for this case is higher than 0.5

Step-by-step explanation:

Data given and notation

n=75 represent the random sample taken

\hat p=0.64 estimated proportion of interest

p_o=0.5 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic

p_v represent the p value

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We want to verify if the true proportion is higher than 0.5:  

Null hypothesis:p =0.5  

Alternative hypothesis:p > 0.5  

The statistic is given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Replacing the info given we got:

z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{75}}}=2.43  

Now we can calculate the p value with the following probability:

p_v =P(z>2.43)=0.0075 \approx 0.008  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion for this case is higher than 0.5

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Step-by-step explanation:

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