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Ivenika [448]
1 year ago
10

A random sample of BYU-Idaho students was surveyed and asked if they were in favor of retaining the penny as a form of currency

in the United States. Out of the 116 women surveyed, 80 said that they were in favor of retaining the penny as a form of currency. Of the 137 men surveyed, 91 said that they were in favor of retaining the penny. For these questions, let group 1 represent women and group 2 represent men. Use this information to answer questions 1 through 7.
1. Are the requirements met to create a confidence interval and conduct a hypothesis test?

Yes. The requirements are met for both.

No. The requirements aren't met for a confidence interval or a hypothesis test.

The requirements are only met for creating a confidence interval.

The requirements are only met for conducting a hypothesis test.
Mathematics
1 answer:
castortr0y [4]1 year ago
8 0

Answer:

Step-by-step explanation:

Hello!

The objective is to compare the proportion of male and female students that are in favor of retaining the penny as a form of currency. Two groups of students were surveyed, the study variables are:

X₁: number of female students of the BYU-Idaho out of 116, that are in favor of retaining the penny as a form of currency.

X₂: number of male students of the BYU-Idaho out of 137, that is in favor of retaining the penny as a form of currency.

These two variables are discrete, and they follow the binomial criteria:

The number of observations of the trial is fixed and there are only two possible observations: success: the student is on the favor, failure: the student is against it.

Each observation in the trial is independent, this means that none of the trials will affect the probability of the next trial.

The probability of success in the same from one trial to another.

So we can say that

X₁~Bi(n₁;p₁)

X₂~Bi(n₂;p₂)

To study the difference between both population proportions through a hypothesis test or a Confidence Interval you theed to apply the Central Limit theorem and approximate the distribution of both sample proportions to normal. For that the sample sizes have to be big enough, as a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

n₁ and n₂ ≥ 30, in this case, n₁= 116 and n₂=137

So the conditions are meet to estimate the difference per CI and test it through a hypothesis test.

I hope this helps!

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Answer:

The blood potassium level L such that the probability is only 0.05 that the average of four measurements is less than L is 3.64.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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What is the blood potassium level L such that the probability is only 0.05 that the average of four measurements is less than L?

This is the value of X when Z has a pvalue of 0.05. So it is X when Z = -1.645.

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

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X - 3.8 = -1.645*0.1

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