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anastassius [24]
3 years ago
7

Mr. Smith takes his children and a few of their friends to the movies and buys each of them a ticket. If the admission for the c

hildren is $5.75 each for a total of eight children, about how much should Mr. Smith expect to spend for their tickets?
Mathematics
2 answers:
Georgia [21]3 years ago
7 0

divide 5.75 by 8

the answer is 0.71 approximately


KiRa [710]3 years ago
6 0
Do 5.75x8 which is 46$ in total
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Since our calculated value is outside than our critical value,z_{calc}=-136.438, we have enough evidence to reject the null hypothesis at 5% of significance. So there is not nough evidence to conclude that mean for Ohio it's significantly different than the mean for Georgia.

Step-by-step explanation:

1) Data given and notation  

\bar X_{1}=3.6 represent the mean for Georgia  

\bar X_{2}=4.15 represent the mean for Ohio  

s_{1}=0.03 represent the population standard deviation for Georgia

\sigma_{2}=0.01 represent the population standard deviation for Ohio

n_{1}=60 sample size for Georgia  

n_{2}=80 sample size for Ohio  

z would represent the statistic (variable of interest)

p_v represent the p value  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the means are different for both groups, the system of hypothesis would be:

H0:\mu_{1} = \mu_{2}  

H1:\mu_{1} \neq \mu_{2}  

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z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

3) Calculate the statistic  

We have all in order to replace in formula (1) like this:  

z=\frac{3.6-4.15}{\sqrt{\frac{0.03^2}{60}+\frac{0.01^2}{80}}}=-136.438  

4) Find the critical value

In order to find the critical value we need to take in count that we are conducting a two tailed test, so we are looking on the normal standard distribution a value that accumulates 0.025 of the area on the right and 0.975 of the area on the left. We can us excel or a table to find it, for example the code in Excel is:

"=NORM.INV(1-0.025,0,1)", and we got z_{critical}=\pm 1.96

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Since our calculated value is outside than our critical value,z_{calc}=-136.438, and outside of (-1.96,1.96), we have enough evidence to reject the null hypothesis at 5% of significance. So there is not nough evidence to conclude that mean for Ohio it's significantly different than the mean for Georgia.

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