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Vera_Pavlovna [14]
3 years ago
15

Food inspectors inspect samples of food products to see if they are safe. This can be thought of as a hypothesis test with the f

ollowing hypotheses.
H0: the food is safe
Ha: the food is not safe
The following is an example of what type of error?
The sample suggests that the food is safe, but it actually is not safe.

a. type I
b. type II
c. not an error
Mathematics
1 answer:
Molodets [167]3 years ago
8 0

Answer:

b. type II

Step-by-step explanation:

given that food inspectors  inspect samples of food products to see if they are safe. This can be thought of as a hypothesis test with the following hypotheses.

H0: the food is safe

Ha: the food is not safe

It was concluded from the hypothesis test that the food is safe while it was not actually safe.

This is a case of false acceptance of null hypothesis when it is false.

In hypothesis test, there are two errors. a type I error is the rejection of a true null hypothesis  while a type II error is the non-rejection of a false null hypothesis

So this is type II error because we did not reject a false null hypothesis.

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Question 10 Please Help
svet-max [94.6K]

Answer:

1)· 5x + 2y = 9. First we solve for y. 2y= 9 -5x. y=(9-5x)/2. Now that we have the value of y. We substitute on the original equation and resolve. 5x + 2y = 95x + 2y = 9  5x + 2(9-5x)/2 = 9  5x + 9 - 5x = 9  9 = 9

That would be x = 1  

Now substitute and resolve to find y.  

5(1) + 2y = 9

5 + 2y = 9

2y = 4

y = 2

So our answer x=1 and y = 2. (1,2)

Proof :  

5(1) + 2(2) = 9

5+ 4 = 9

9 = 9

Step-by-step explanation:

hoped it helped for the first  one  i didnt now the second one

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Answer:

(a) See below

(b) r = 0.9879  

(c) y = -12.629 + 0.0654x

(d) See below

(e) No.

Step-by-step explanation:

(a) Plot the data

I used Excel to plot your data and got the graph in Fig 1 below.

(b) Correlation coefficient

One formula for the correlation coefficient is  

r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}

The calculation is not difficult, but it is tedious.

(i) Calculate the intermediate numbers

We can display them in a table.

<u>    x   </u>    <u>      y     </u>   <u>       xy     </u>    <u>              x²    </u>   <u>       y²    </u>

   36       0.22              7.92               1296           0.05

   67        0.62            42.21              4489           0.40

   93         1.00            93.00           20164           3.46

 433        11.8          5699.4          233289        139.24

 887      29.3         25989.1          786769       858.49

1785      82.0        146370          3186225      6724

2797     163.0         455911         7823209    26569

<u>3675 </u>  <u> 248.0  </u>    <u>   911400      </u>  <u>13505625</u>   <u> 61504        </u>

9965   537.81     1545776.75  25569715   95799.63

(ii) Calculate the correlation coefficient

r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}\\\\= \dfrac{9\times 1545776.75 - 9965\times 537.81}{\sqrt{[9\times 25569715 -9965^{2}][9\times 95799.63 - 537.81^{2}]}} \approx \mathbf{0.9879}

(c) Regression line

The equation for the regression line is

y = a + bx where

a = \dfrac{\sum y \sum x^{2} - \sum x \sum xy}{n\sum x^{2}- \left (\sum x\right )^{2}}\\\\= \dfrac{537.81\times 25569715 - 9965 \times 1545776.75}{9\times 25569715 - 9965^{2}} \approx \mathbf{-12.629}\\\\b = \dfrac{n \sum xy  - \sum x \sum y}{n\sum x^{2}- \left (\sum x\right )^{2}} -  \dfrac{9\times 1545776.75  - 9965 \times 537.81}{9\times 25569715 - 9965^{2}} \approx\mathbf{0.0654}\\\\\\\text{The equation for the regression line is $\large \boxed{\mathbf{y = -12.629 + 0.0654x}}$}

(d) Residuals

Insert the values of x into the regression equation to get the estimated values of y.

Then take the difference between the actual and estimated values to get the residuals.

<u>    x    </u>   <u>      y     </u>   <u>Estimated</u>   <u>Residual </u>

    36        0.22        -10                 10

    67        0.62          -8                  9

    93        1.00           -7                  8

   142        1.86           -3                  5

  433       11.8             19               -  7

  887     29.3             45               -16  

 1785     82.0            104              -22

2797    163.0            170               -  7

3675   248.0            228               20

(e) Suitability of regression line

A linear model would have the residuals scattered randomly above and below a horizontal line.

Instead, they appear to lie along a parabola (Fig. 2).

This suggests that linear regression is not a good model for the data.

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