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Svetradugi [14.3K]
3 years ago
5

The most recent public health statistics available indicate that 23.6​% of American adults smoke cigarettes. Using the​ 68-95-99

.7 Rule, describe the sampling distribution model for the proportion of smokers among a randomly selected group of 40 adults. Be sure to discuss your assumptions and conditions. Describe the sampling distribution. There is a​ 68% chance that between nothing​% and nothing​% are​ smokers, a​ 95% chance that between nothing​% and nothing​% are​ smokers, and a​ 99.7% chance that between nothing​% and nothing​% are smokers. ​(Round to one decimal place as​ needed.)
Mathematics
1 answer:
artcher [175]3 years ago
7 0

Answer:

There is a​ 68% chance that between 17​% and 30​% are​ smokers.

There is a​ 95% chance that between 10​% and 37​% are​ smokers.

There is a​ 99.7% chance that between 4​% and 44​% are​ smokers.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes <em>n</em> > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of the sampling distribution of sample proportion is:

 \mu_{\hat p}=p\\

The standard deviation of the sampling distribution of sample proportion is:

 \sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}

Given:

<em>n</em> = 40

<em>p</em> = 0.236

Compute the mean and standard deviation of this sampling distribution of sample proportion as follows:

\mu_{\hat p}=p=0.236

\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.236(1-0.236)}{40}}=0.067

The Empirical Rule states that in a normal distribution with mean <em>µ</em> and standard deviation <em>σ</em>, nearly all the data will fall within 3 standard deviations of the mean. The empirical rule can be divided into three parts:

  • 68% data falls within 1 standard-deviation of the mean.  

        That is P (µ - σ ≤ X ≤ µ + σ) = 0.68.

  • 95% data falls within 2 standard-deviations of the mean.

        That is P (µ - 2σ ≤ X ≤ µ + 2σ) = 0.95.

  • 99.7% data falls within 3 standard-deviations of the mean.

        That is P (µ - 3σ ≤ X ≤ µ + 3σ) = 0.997.

Compute the range of values that has a probability of 68% as follows:

P (\mu_{\hat p} - \sigma_{\hat p} \leq  \hat p \leq  \mu_{\hat p} + \sigma_{\hat p}) = 0.68\\P(0.236-0.067\leq  \hat p \leq 0.236+0.067)=0.68\\P(0.169\leq  \hat p \leq0.303)=0.68\\P(0.17\leq  \hat p \leq0.30)=0.68

Thus, there is a​ 68% chance that between 17​% and 30​% are​ smokers.

Compute the range of values that has a probability of 95% as follows:

P (\mu_{\hat p} - 2\sigma_{\hat p} \leq  \hat p \leq  \mu_{\hat p} + 2\sigma_{\hat p}) = 0.95\\P(0.236-2\times 0.067\leq  \hat p \leq 0.236+2\times0.067)=0.95\\P(0.102\leq  \hat p \leq 0.370)=0.95\\P(0.10\leq  \hat p \leq0.37)=0.95

Thus, there is a​ 95% chance that between 10​% and 37​% are​ smokers.

Compute the range of values that has a probability of 99.7% as follows:

P (\mu_{\hat p} - 3\sigma_{\hat p} \leq  \hat p \leq  \mu_{\hat p} + 3\sigma_{\hat p}) = 0.997\\P(0.236-3\times 0.067\leq  \hat p \leq 0.236+3\times0.067)=0.997\\P(0.035\leq  \hat p \leq 0.437)=0.997\\P(0.04\leq  \hat p \leq0.44)=0.997

Thus, there is a​ 99.7% chance that between 4​% and 44​% are​ smokers.

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