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poizon [28]
3 years ago
13

Data from a sample of citizens of a certain country yielded the following estimates of average TV viewing time per month for all

the citizens. The times are in hours and minutes.​ (NA, not​ available)
Viewing Method May 2008 May 2007 Change (%)
Watching TV in the home

Watching timeshifted TV

Using the internet

Watching video on internet

127:49

5:31

26:41

2:34

121:59

3:52

24:17

NA

5

43

10

NA

Is the study descriptive or​ inferential?

A. ​Inferential, because the statistics are used to describe the sample

B. ​Descriptive, because the statistics are used to describe the sample

C. ​Descriptive, because the statistics are used to make an inference about the population

D. ​Inferential, because the statistics are used to make an inference about the population
Mathematics
1 answer:
Alexeev081 [22]3 years ago
8 0

Answer:

D. ​Inferential, because the statistics are used to make an inference about the population

Correct, the objective of this study is obtain information from the population with a sample and then use any method to estimate the population mean, the parameter of interest.

Step-by-step explanation:

An inferential study consists in take information about a population by a sample and use this information to see what would be the possible values for the population of interest

By the other hand a descriptive study is obtained from observing and measuring some variables of interest but without manipulate the data.

For this case we have sample averages for the viewing time per month.

Let's analyze one by one the possible options:

A. ​Inferential, because the statistics are used to describe the sample

False the study is inferential but the idea is not just obtain information about the sample, we want to see the population parameters not the statistics

B. ​Descriptive, because the statistics are used to describe the sample

False for this case we have averages calculated from the sample mean and is not possible to consider this study as descriptive.

C. ​Descriptive, because the statistics are used to make an inference about the population

False, the statistics are used to make an inference about the population, this statement is correct, but the problem is that this study is not descriptive.

D. ​Inferential, because the statistics are used to make an inference about the population

Correct, the objective of this study is obtain information from the population with a sample and then use any method to estimate the population mean, the parameter of interest.

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Suppose that 20% of the residents in a certain state support an increase in the property tax. An opinion poll will randomly samp
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Answer:

95.44% probability the resulting sample proportion is within .04 of the true proportion.

Step-by-step explanation:

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

Normal probability distribution

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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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the sampling distribution of the sample proportion in sample of size n, the mean is \mu = p and the standard deviation is s = \sqrt{\frac{p(1-p)}{n}}

In this question:

p = 0.2, n = 400

So

\mu = 0.2, s = \sqrt{\frac{0.2*0.8}{400}} = 0.02

How likely is the resulting sample proportion to be within .04 of the true proportion (i.e., between .16 and .24)?

This is the pvalue of Z when X = 0.24 subtracted by the pvalue of Z when X = 0.16.

X = 0.24

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

By the Central Limit Theorem

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

Z = \frac{0.24 - 0.2}{0.02}

Z = 2

Z = 2 has a pvalue of 0.9772.

X = 0.16

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

Z = \frac{0.16 - 0.2}{0.02}

Z = -2

Z = -2 has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

95.44% probability the resulting sample proportion is within .04 of the true proportion.

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