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ElenaW [278]
3 years ago
12

675 students in a school .canteen is giving 100 free apples for all the students .calculate how much would a child receive from

the apple
Mathematics
2 answers:
Nina [5.8K]3 years ago
5 0

Since there are 675 students in a school and the canteen is giving 100 free apples to all the students, therefore, one student will receive less than one apple.

One student will receive \frac{100}{675}=\frac{4}{27} part of an apple.

This information can be further interpreted as a percentage by doing the following calculation:

\frac{4}{27}\times 100\approx14.81%

Thus, we can say that one student receives 14.81% of an apple. (Note that 100% represents one full apple).

Inessa05 [86]3 years ago
4 0
That would be 100/675 =  4/27 of an apple
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The Salk polio vaccine experiment in 1954 focused on the effectiveness of the vaccine in combating paralytic polio. Because it w
sdas [7]

Answer:

Step-by-step explanation:

Hello!

The variables of interest are:

X₁: Number of cases of polio observed in kids that received the placebo vaccine.

n₁= 201299 total children studied

x₁= 110 cases observed

X₂: Number of cases of polio observed in kids that received the experimental vaccine.

n₂= 200745 total children studied

x₂= 33 cases observed

These two variables have a binomial distribution. The parameters of interest, the ones to compare, are the population proportions: p₁ vs p₂

You have to test if the population proportions of children who contracted polio in both groups are different: p₂ ≠ p₁

a)

H₀: p₂ = p₁

H₁: p₂ ≠ p₁

α: 0.05

Z= \frac{(p'_2-p'_1)-(p_2-p_1)}{\sqrt{p'[\frac{1}{n_1} +\frac{1}{n_2} ]} }

Sample proportion placebo p'₁= x₁/n₁= 110/201299= 0.0005

Sample proportion vaccine p'₂= x₂/n₂= 33/200745= 0.0002

Pooled sample proportion p'= (x₁+x₂)/(n₁+n₂)= (110+33)/(201299+200745)= 0.0004

Z_{H_0}= \frac{(0.0002-0.0005)-0}{\sqrt{0.0004[\frac{1}{201299} +\frac{1}{200745} ]} }= -4.76

This test is two-tailed, using the critical value approach, you have to determine two critical values:

Z_{\alpha/2}= Z_{0.025}= -1.96

Z_{1-\alpha /2}= Z_{0.975}= 1.96

Then if Z_{H_0} ≤ -1.96 or if Z_{H_0} ≥ 1.96, the decision is to reject the null hypothesis.

If -1.96 < Z_{H_0} < 1.96, the decision is to not reject the null hypothesis.

⇒ Z_{H_0}= -4.76, the decision is to reject the null hypothesis.

b)

H₀: p₂ = p₁

H₁: p₂ ≠ p₁

α: 0.01

Z= \frac{(p'_2-p'_1)-(p_2-p_1)}{\sqrt{p'[\frac{1}{n_1} +\frac{1}{n_2} ]} }

The value of Z_{H_0}= -4.76 doesn't change, since we are working with the same samples.

The only thing that changes alongside with the level of significance is the rejection region:

Z_{\alpha /2}= Z_{0.005}= -2.576

Z_{1-\alpha /2}= Z_{0.995}= 2.576

Then if Z_{H_0} ≤ -2.576or if Z_{H_0} ≥ 2.576, the decision is to reject the null hypothesis.

If -2.576< Z_{H_0} < 2.576, the decision is to not reject the null hypothesis.

⇒ Z_{H_0}= -4.76, the decision is to reject the null hypothesis.

c)

Remember the level of significance (probability of committing type I error) is the probability of rejecting a true null hypothesis. This means that the smaller this value is, the fewer chances you have of discarding the true null hypothesis. But as you know, you cannot just reduce this value to zero because, the smaller α is, the bigger β (probability of committing type II error) becomes.

Rejecting the null hypothesis using different values of α means that there is a high chance that you reached a correct decision (rejecting a false null hypothesis)

I hope this helps!

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Answer:

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