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ELEN [110]
3 years ago
12

In order to estimate the average time spent on the computer terminals per student at a local university, data were collected fro

m a sample of 81 business students over a one-week period. Assume the population standard deviation is 1.2 hours. Calculate the margin of error (to two decimals) for a confidence interval with a confidence coefficient of 0.95.
Mathematics
1 answer:
VladimirAG [237]3 years ago
4 0

Answer:

Margin of error  for a 95% of confidence intervals is 0.261

Step-by-step explanation:

<u>Step1:-</u>

 Sample n = 81 business students over a one-week period.

 Given the population standard deviation is 1.2 hours

 Confidence level of significance = 0.95

 Zₐ = 1.96

Margin of error (M.E) = \frac{Z_{\alpha  }S.D }{\sqrt{n} }

Given n=81 , σ =1.2 and  Zₐ = 1.96

<u>Step2:-</u>

<u />Margin of error (M.E) = \frac{Z_{\alpha  }S.D }{\sqrt{n} }<u />

<u />Margin of error (M.E) = \frac{1.96(1.2) }{\sqrt{81} }<u />

On calculating , we get

Margin of error = 0.261

<u>Conclusion:-</u>

Margin of error  for a 95% of confidence intervals is 0.261

<u />

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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:

Option B.

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