Please help me brainlessly

nekit [7.7K]

6/11 is a simplified version of the fraction and its the most simple you can get.

5 0

3 years ago

Gun ownership Concerned about recent incidence of gun violence, a public interest group conducts a poll of 850 randomly selected

denis23 [38]

Answer:

a.

The 95% confidence interval for the proportion of all American adults who have a gun in their home is (0.4066, 0.4734). This means that we are 95% sure that the true population proportions of American adults who have a gun in their home is between these two values.

b.

The confidence level is the probability of the confidence interval containing the population proportion.

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Poll of 850 randomly selected American adults and finds that 44% of those surveyed have a gun in their home.

This means that $n = 850, \pi = 0.44$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.44 - 1.96\sqrt{\frac{0.44*0.56}{850}} = 0.4066$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.44 + 1.96\sqrt{\frac{0.44*0.56}{850}} = 0.4734$

The 95% confidence interval for the proportion of all American adults who have a gun in their home is (0.4066, 0.4734). This means that we are 95% sure that the true population proportions of American adults who have a gun in their home is between these two values.

b. Explain what is meant by 95% confidence in this context.

The confidence level is the probability of the confidence interval containing the population proportion.

5 0

3 years ago

According to a 2009 Reader's Digest article, people throw away approximately 11% of what they buy at the grocery store. Assume t

expeople1 [14]

Answer:

0.14% probability that the sample proportion exceeds 0.2

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

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

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .