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labwork [276]
3 years ago
13

Anita's, a fast-food chain specializing in hot dogs and garlic fries, keeps track of the proportion of its customers who decide

to eat in the restaurant (as opposed to ordering the food "to go"), so it can make decisions regarding the possible construction of in-store play areas, the attendance of its mascot Sammy at the franchise locations, and so on. Anita's reports that of its customers order their food to go. If this proportion is correct, what is the probability that, in a random sample of customers at Anita's, exactly order their food to go?
Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
8 0

Answer: Pt = Ng / Nt

Step-by-step explanation:

Where Ng = number of customers that order food to go as reported at Anita's

Nt = total number of customers reported at Anita's

Pt = probability that a customer will order food to go at Anita's

Pt= Ng/Nt

Goodluck...

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GREYUIT [131]

Answer:

z = \frac{0.00206 -0.0110}{\sqrt{\frac{0.00206*(1-0.00206)}{7765} +\frac{0.0110*(1-0.0110)}{2823}}}= -4.405

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Since the p value is very low compared to the significance level of 0.05 we have enough evidence to reject the null hypothesis and then the best conclusion would be:

B) Since p <a, we conclude that this data shows that seat belts are effective in reducing fatalities.  

Step-by-step explanation:

We have the following info given from the problem

X_1 =31 number of people killed with occupants not wering seat belts

n_1 = 2823 number of people not wearing seat belts

\hat p_1 =\frac{31}{2823}=0.0110 represent the estimated proportion of people killed not wearing seatbelts

X_2 =16 number of people killed with occupants using wering seat belts

n_2 = 7765 number of people using wearing seat belts

\hat p_2 =\frac{16}{7765}=0.00206 represent the estimated proportion of people killed  wearing seatbelts

We want to check if seat belts are effective in reducing fatalities, so we want to test the following system of hypothesis:

Null hypothesis: p_2 \geq p_1

Alternative hypothesis: p_2

The statistic for this case is given by:

z = \frac{\hat p_2 -\hat p_1}{\sqrt{\frac{\hat p_1 (1-\hat p_1)}{n_1} +\hat p_2 (1-\hat p_2)}{n_2}}}

And replacing we got:

z = \frac{0.00206 -0.0110}{\sqrt{\frac{0.00206*(1-0.00206)}{7765} +\frac{0.0110*(1-0.0110)}{2823}}}= -4.405

Now we can find the p value since we are using a left tailed test the p value would be:

p_v = P(z

Since the p value is very low compared to the significance level of 0.05 we have enough evidence to reject the null hypothesis and then the best conclusion would be:

B) Since p <a, we conclude that this data shows that seat belts are effective in reducing fatalities.  

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We know that

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Answer:

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25=c

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