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Wittaler [7]
3 years ago
8

A florist wants to determine if a new additive would extend the life of cut flowers longer than the original additive. The flori

st randomly selects 20 carnations from the ones recently delivered by the greenhouse and places the first 10 in water with the new additive and remaining 10 in water with the original additive. After three weeks, 6 carnations placed in the new additive still looked healthy and 2 carnations placed in the original additive still looked healthy. The proportion of healthy carnations with the new additive was significantly greater than the proportion of healthy carnations with the original additive.
Which of the following is a valid conclusion?

Conclusions about cause and effect for the additives can be made, because the florist randomly selected the 20 carnations; additionally, inferences can be made about the population of carnations at the greenhouse.
Conclusions about cause and effect for the additives cannot be made, because the florist did not randomly assign the 20 carnations; however, inferences can be made about the population of carnations at the greenhouse, because the sample was random.
Conclusions about cause and effect for the additives can be made, because the florist took a random sample of 20 carnations; however, inferences cannot be made about the population of carnations at the greenhouse, because the carnations were not randomly assigned to the treatments.
Conclusions about cause and effect for the additives cannot be made, because the florist took a random sample of 20 carnations; and, inferences cannot be made about the population of carnations at the greenhouse, because the carnations were not randomly assigned to the treatments.
Mathematics
1 answer:
bekas [8.4K]3 years ago
7 0

Answer:

The last one would be a valid conclusion {Conclusions about cause and effect for the additives cannot be made, because the florist took a random sample of 20 carnations; and, inferences cannot be made about the population of carnations at the greenhouse, because the carnations were not randomly assigned to the treatments.}

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Answer:

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And the deviation would be:

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Step-by-step explanation:

For this case we have the following distribution given:

X        0         1       2       3       4         5        6

P(X)  0.3   0.25   0.2   0.12   0.07   0.04   0.02

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Now we can find the second moment given by:

E(X^2) =\sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

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And the variance would be given by:

Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779

And the deviation would be:

Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54

 

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