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N76 [4]
4 years ago
11

Why aren’t voluntary response samples usually representative of the population ?

Mathematics
2 answers:
FromTheMoon [43]4 years ago
7 0

The second choice is correct. An example of this is a voluntary "product feedback surveys" which tend to biased toward the negative. While people with both positive and negative feedback respond overall, customers who have a complaint are more motivated to take the time, and thus are overrepresented.

stira [4]4 years ago
4 0

definitely the second, the people who volunteer usually have a motive to do so and will be the only ones represented.

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The ratio of frogs to swans is 2:6. What is the ratio of swans to the total number of animals.
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4 years ago
Read 2 more answers
A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics e
professor190 [17]

Answer:

\chi^2 =\frac{15-1}{25} 16 =8.96

The degrees of freedom are given by:

df = n-1 = 15-1=14

The p value for this case would be given by:

p_v =P(\chi^2

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

Step-by-step explanation:

Information given

n=15 represent the sample size

\alpha=0.05 represent the confidence level  

s^2 =16 represent the sample variance

\sigma^2_0 =25 represent the value that we want to  verify

System of hypothesis

We want to test if the true deviation for this case is lesss than 5minutes, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 \geq 25

Alternative hypothesis: \sigma^2

The statistic is given by:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

And replacing we got:

\chi^2 =\frac{15-1}{25} 16 =8.96

The degrees of freedom are given by:

df = n-1 = 15-1=14

The p value for this case would be given by:

p_v =P(\chi^2

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

6 0
3 years ago
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