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Mice21 [21]
3 years ago
13

A potato chip company produces a large number of potato chip bags each day and wants to investigate whether a new packaging mach

ine will lower the proportion of bags that are damaged. The company selected a random sample of 150 bags from the old machine and found that 15 percent of the bags were damaged, then selected a random sample of 200 bags from the new machine and found that 8 percent were damaged. Let p^O represent the sample proportion of bags packaged on the old machine that are damaged, pˆN represent the sample proportion of bags packaged on the new machine that are damaged, pˆC represent the combined proportion of damaged bags from both machines, and nO and nN represent the respective sample sizes for the old machine and new machine.
Required:
Have the conditions for statistical inference for testing a difference in population proportions been met?
Mathematics
1 answer:
alexandr1967 [171]3 years ago
4 0

Answer:

The conditions for the approximation have been met

Step-by-step explanation:

Random samples on both cases  and

p∧o proportion of bags packaged on the old machine that are damaged

p∧n sample proportion of bags packaged on the new machine that are damaged

p∧c  =  ( p∧o + p∧n ) / (nO + nN )  

p∧o = 15 %      p∧o =  0,15         nO = 150

p∧o *  nO = 0,15 *150          p∧o *  nO = 22,5

p∧n  = 8 %      p∧n  = 0,08     nN = 200

p∧n*nN  = 0,08 * 200     p∧n*nN  = 16

Both  p∧n*nN  and   p∧o *  nO    are bigger than 5

Therefore both samples are big enough for the approximation of the binomial distribution to normal distribution

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Answer:

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Step-by-step explanation:

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3 years ago
A single card is drawn from a standard deck of cards. Find the probability of drawing a club or a diamond.
Akimi4 [234]

Answer:

1/10

Step-by-step explanation:

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2 years ago
I NEED HELP ASAP!!<br><br> 4(5)(-2)=?
vova2212 [387]
So, in an easier way to say it
4 × 5 × -2

You would start by doing 
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5 0
3 years ago
Assume that a cross is made between a heterozygous tall pea plant and a homozygous short pea plant. Fifty offspring are produced
Butoxors [25]

Answer:

1) \chi^2 = \frac{(30-25)^2}{25} +\frac{(20-25)^2}{25} =1 +1 =2

2) df = c-1 = 2-1

Where c represent the number of categories c=2

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Tall =30 , Short =20

We need to conduct a chi square test in order to check the following hypothesis:

H0: The deviations from a 1:1 ratio (25 tall and 25 short) are due to chance.

H1: The deviations from a 1:1 ratio (25 tall and 25 short) are NOT due to chance.

Part 1

So then we know that the expected values would be 25 for each case

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

And if we replace we got:

\chi^2 = \frac{(30-25)^2}{25} +\frac{(20-25)^2}{25} =1 +1 =2

Part 2

For this case the degreed of freedom are given by:

df = c-1 = 2-1

Where c represent the number of categories c=2

And we can calculate the p value given by:

p_v = P(\chi^2_{1} >2)=0.157

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(2,1,TRUE)"

7 0
3 years ago
Please HeLP me por favor​
miv72 [106K]

Answer:

A: x<3 (x is smaller than three)

6 0
2 years ago
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