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vitfil [10]
2 years ago
8

You are thinking about opening a restaurant and are searching for a good location. From research you have done, you know that th

e mean income of those living near the restaurant must be over $85,000 to support the type of upscale restaurant you wish to open. You decide to take a simple random sample of 50 people living near one potential location. Based on the mean income of this sample, you will decide whether to open a restaurant there. Describe a Type I and a Type II error, and explain the consequences of each
Mathematics
1 answer:
harina [27]2 years ago
6 0

The type I error and type II error can be described as follows:

Type I error: We can conclude that if the mean exceeds $85000, when in fact it does not. Thus, opening your restaurant in a locale that will not support it).

Type II error: We can conclude that if the mean income does not exceed $85000, when in fact it does. Thus, deciding not to open your restaurant in a locale that will support it.

<h3>What is a Type I and a Type II error?</h3>

A Type I error in statistics is described as rejecting the null hypothesis when it is actually true, and a Type II error is defined as failing to reject the null hypothesis when it is genuinely untrue.

The type I error and type II error can be described as follows:

  • Type I error: We can conclude that if the mean exceeds $85000, when in fact it does not.
  • Type II error: We can conclude that if the mean income does not exceed $85000, when in fact it does.

The consequences of each error are:

If you launched your business in an inappropriate location, you would incur a financial loss before realizing your error.

If you do not open your restaurant in an ideal location, you will miss out on the potential to make a profit, but you will not necessarily lose money.

Learn more about type I error and type II error here:

brainly.com/question/16012410

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Paha777 [63]

Answer:

(1\frac{1}{15},3\frac{5}{12})

Step-by-step explanation:

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To find an average of a data with 2 elements you must add the two elements then take that sum and divide it by 2.

So this is what we will do with the x's then the y's.

So first x:

\frac{4\frac{1}{3}+-2\frac{1}{5}}{2}

Write as improper fractions:

\frac{\frac{13}{3}+\frac{-11}{5}}{2}

Division by 2 is the same as multiplying by 1/2:

\frac{13}{6}+\frac{-11}{10}

The least common multiple of 6 and 10 is 30.  We will multiply first fraction by 5/5 and 3/3 for the second so we can have the same denominator.

\frac{65}{30}+\frac{-33}{30}

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\frac{32}{30}

If you want the answer as a mix fraction like your question began as we can do that by first figuring out how many 30's are in 32 and what is the remainder of that division.

There is one 30 and 32 and so 32-30=2 is the remainder.

1\frac{2}{30}

Reduce by dividing top and bottom by 2:

1\frac{1}{15}

Now for y:

\frac{3\frac{1}{6}+3\frac{2}{3}}{2}

Write the mix fractions as improper fractions:

\frac{\frac{19}{6}+\frac{11}{3}}{2}

Dividing by 2 is the same as multiplying by 1/2:

\frac{19}{12}+\frac{11}{6}

The least common multiple of 12 and 6 is 12 so I'm going to multiply the second fraction by 2/2 so the denominators will be the same:

\frac{19}{12}+\frac{22}{12}

Since the denominators are the same we can write as a single fraction:

\frac{41}{12}

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3\frac{5}{12}

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(1\frac{1}{15},3\frac{5}{12}).

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3 years ago
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Since angle KLJ~angle VWU are similar :

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