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alexdok [17]
2 years ago
5

A company rents out 11 food booths and 24 game booths at the county fair. The fee for a food booth is $175 plus $5 per day. The

fee for a game booth is $50 plus $7 per day. The fair lasts for d days, and all the booths are rented for the entire time. Enter a simplified expression for the amount, in dollars, that the company is paid.
The expression is ...
Mathematics
1 answer:
AURORKA [14]2 years ago
3 0
11(175+5d) + 24(50+7d)
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Answer:

1) Null hypothesis:\mu \leq 500  

Alternative hypothesis:\mu > 500  

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Step-by-step explanation:

Part 1

Data given

\bar X=640 represent the sample mean

\sigma=150 represent the population standard deviation

n=40 sample size  

\mu_o =500 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)  

Step1:State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is higher than 500, the system of hypothesis would be:  

Null hypothesis:\mu \leq 500  

Alternative hypothesis:\mu > 500  

Step 2: Calculate the statistic

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

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

z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90  

Step 3: Calculate the critical value

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

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Step 4: Compare the statistic with the critical value

For this case we see that the calculated value is higher than the critical value

Step 5: Decision

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

Part 2

P-value  

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p_v =P(z>5.90)=1.82x10^{-9}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

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