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lawyer [7]
4 years ago
5

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been deter

mined that as long as the mean temperature of the discharged water is at most 150°F, there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge water temperature above 150°, 50 water samples will be taken at randomly selected times and the temperature of each sample recorded. The resulting data will be used to test the hypotheses H0: μ = 150° versus Ha: μ > 150°. In the context of this situation, describe type I error.
Mathematics
1 answer:
blondinia [14]4 years ago
8 0

Answer:

The system of hypothesis for this case are:

Null hypothesis: \mu =150

Alternative hypothesis: \mu >150

A type of error I for this case would be reject the null hypothesis that the true mean is equal than 150 when actually is not true.

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".  

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".  

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.  

Type II error, also known as a "false negative" is the error of not rejecting a null  hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.  

Solution to the problem

The system of hypothesis for this case are:

Null hypothesis: \mu =150

Alternative hypothesis: \mu >150

A type of error I for this case would be reject the null hypothesis that the true mean is equal than 150 when actually is not true.

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One troop member So tickets for 33 crab meals and 20 vegetarian meals with a total receipt of $2025 another sold tickets for 30
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Answer:

Therefore the cost of a crab meal and a vegetarian are $40.41 and $1.94 respectively.

Step-by-step explanation:

Given that,tickets for 33 crab meals and 20 vegetarian meal with total receipt of $2025. Another sold ticket for 30 crab meal and 11 vegetarian meals with total receipt of $1827.

Let the cost of a crab meal and a vegetarian meal be x and y respectively.

The cost of a crab meal is x

The cost of 33 crab meal is $33x

The cost of 30 crab meal is $30x

The cost of a vegetarian meal is y.

The cost of 20 vegetarian meal is $20y.

The cost of 11 vegetarian meal is $11y

According to problem,

33x+20y=2025

33x+20y-2025=0........(1)

        and

30x+11y=1827

30x+11y-1827=0 ..........(2)

Cross Multiplication Method,

a_1x+b_1y+c_1=0

a_2x+b_2y+c_2=0

\frac{x}{b_1c_2-b_2c_2}=\frac{y}{a_2c_1-a_1c_2}=\frac{1}{a_1b_2-a_2b_1}

{x}=\frac{b_1c_2-b_2c_2}{a_1b_2-a_2b_1}

{y}=\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}

Here a_1=33,b_1=20,c_1=-2025, a_2=30,b_2=11,c_2=-1827

\therefore x=\frac{20.(-1827)-11.(-2025)}{33.11-30.20}                   y=\frac{30.(-2025)-33.(-1827)}{33.11-30.20}

     \approx 40.41                                              \approx 1.94

Therefore the cost of a crab meal and a vegetarian are $40.41 and $1.94 respectively.

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