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Vilka [71]
3 years ago
8

A researcher wants to survey people from different age groups to analyze voting patterns. Which type of survey will best accompl

ish this goal?
systematic random sampling
voluntary sampling
convenience sampling
stratified sampling
Mathematics
2 answers:
nordsb [41]3 years ago
6 0

Answer: Stratified sampling


Step-by-step explanation:

Stratified sampling is a random sampling method in which the population is divided into non-coinciding groups are called strata and a sample is picked out by some design within each stratum.

Given: A researcher wants to survey people from different age groups to analyze voting patterns.

Thus he will make different groups according to different ages to analyze voting patterns that are called stratum.

Thus, Stratified sampling best accomplish this goal.


tiny-mole [99]3 years ago
4 0

This is an example of stratified sampling because here the samples are divided into different age groups. A stratified sampling is a better way of sampling as it provides a greater precision of results out of a small sample size. This method is also economic as it operates on less samples.

Hence, stratified sampling is the correct answer.

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Mattias bought a package of 5 notebooks for $13.95. What is the unit price per notebook?
Marysya12 [62]

Answer:

2.79

Step-by-step explanation:

this solution:

5x=13.95

x=13.95/5

x2.79

so answer is 2.79

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In a recent​ year, a poll asked 2362 random adult citizens of a large country how they rated economic conditions. In the​ poll,
Harman [31]

Answer:

a) The 99% confidence interval is given by (0.198;0.242).

b) Based on the p value obtained and using the significance level assumed \alpha=0.01 we have p_v>\alpha so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.

c) \alpha=0.01

Step-by-step explanation:

<em>Data given and notation   </em>

n=2362 represent the random sample taken

X represent the people who says that  they would watch one of the television shows.

\hat p=\frac{X}{n}=0.22 estimated proportion of people rated as​ Excellent/Good economic conditions.

p_o=0.24 is the value that we want to test

\alpha represent the significance level  

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  <em> </em>

<em>Concepts and formulas to use   </em>

We need to conduct a hypothesis in order to test the claim that 24% of people are rated with good economic conditions:  

Null hypothesis:p=0.24  

Alternative hypothesis:p \neq 0.24  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

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

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Part a: Test the hypothesis

<em>Check for the assumptions that he sample must satisfy in order to apply the test   </em>

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.  

b) The sample needs to be large enough

np = 2362x0.22=519.64>10 and n(1-p)=2364*(1-0.22)=1843.92>10

Condition satisfied.

<em>Calculate the statistic</em>  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.22 -0.24}{\sqrt{\frac{0.24(1-0.24)}{2362}}}=-2.28

The confidence interval would be given by:

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

The critical value using \alpha=0.01 and \alpha/2 =0.005 would be z_{\alpha/2}=2.58. Replacing the values given we have:

0.22 - (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.198

 0.22 + (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.242

So the 99% confidence interval is given by (0.198;0.242).

Part b

<em>Statistical decision   </em>

P value method or p value approach . "This method consists on determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided is \alpha=0.01. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z  

So based on the p value obtained and using the significance level assumed \alpha=0.01 we have p_v>\alpha so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.

Part c

The confidence level assumed was 99%, so then the signficance is given by \alpha=1-confidence=1-0.99=0.01

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Answer:

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Step-by-step explanation:

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Answer:

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