Question

A study was conducted to determine the proportion of American teenagers between 13 and 17 who smoke. Previous surveys showed that 15% percent of all teenagers smoke. A Gallup survey interviewed a nationally representative sample of 785 teenagers aged 13 to 17. Seventy-one teenagers in the survey acknowledged having smoked at least once in the past week. Does the study provide adequate evidence to conclude that the percentage of teenagers who smoke is now different than 15%? Be sure to answer all the questions whether or not the requirements are actually met.

Answer 1

Answer:

The study only provides evidence that the percentage of teenagers who smoke is different than 15% but only considering teenagers between 13 and 17.

Step-by-step explanation:

I will assume that when we talk about a teenager, we are talking about a teenager between 13 and 17 years old. We can solve this problem with a hypothesis test. Lets first define the main hypothesis of our test and the alternate hypothesis, note that 70 is way less than 15% of 785 (is less than 10%), thus we can use the following ones

H0: 15% of the teenagers smoke

H1: Less than 15% of teenagers smoke

Lets rewrite H0 and H1 using probabilities. Let X be the amount of teenagers that smoke in a sample with length 785. X is a binomial random variable. If we take H0 to be true, then the probability of success in each individual outcome of X is 0.15. This means that the mean is μ = 0.15*785 = 117.75, and the standard deviation is σ = √(117.75(1-0.15)) = 10.00437.

Since we are working with a sample of length high enough (> 30), then the Central Limit Theorem tells us that X behives pretty similar to a Normal random variable, with similar mean and standard deviation; therefore, we may assume directly that X is normal.

The hypothesis can be rewritten in terms of X this way:

H0: μ = 117.75 (this means that in average 15% of a sample of 785 smoke)

H1: μ < 117.75

We will use a 95% confidence interval. note that if only 70 teenagers smoke, then that means that SX = 70, where SX is the sample we obtain. We will calculate the probability that X is less than (or equal) to 70, if that probability is less than 0.05, then we can say that we have evidence that the percentage of teenagers who smoke is different (in fact, less), than 15%.

In order to calculate P(X < 70), we will use the standarization of X, given by

$W = \frac{X-\mu}{\sigma} = \frac{X-117.75}{10.00437}$

The cummulative distribution function of W, which we denote $\phi$ has well known values and they can be found in the attached file. Also, since the density function of a standard random normal variable is symmetric, then we have that $\phi(-x) = 1-\phi(x)$ for any positive value x.

$P(X < 70) = P(\frac{X-117.75}{10.00437} < \frac{70-117.75}{10.00437}) = P(W< -4.772914) = \phi(-4.772914) = 1-\phi(4.772914)$

If we look at the table, we will realise that $\phi(4.772914)$ is practically 1, thus P(X < 70) is practically 0 if we assume that the mean of X is 117.75.

This means that we have evidence that the percentage of teenagers who smoke is no longer 15%, it is less.

However, here we are assuming that the term 'teenager' and teenager between 13 and 17' is the same. Maybe the the study took teenagers with 20 years old or so, and if that happened, then it makes sense that the results here are not the same. Therefore, we conclude that the study only provides evidence that the percentage of teenagers who smoke is different than 15% but only considering teenagers between 13 and 17 (because that is where the sample came from).

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