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Eva8 [605]
3 years ago
11

Which of the following is not true about cluster sampling?

Mathematics
2 answers:
Step2247 [10]3 years ago
8 0

Answer:

d

Step-by-step explanation:

just took it

Aloiza [94]3 years ago
5 0

Answer:

The correct answer is D. It is not true that cluster sampling uses randomly selected clusters and samples everyone within each cluster.

Step-by-step explanation:

Cluster sampling is a method of collecting samples and statistical data, by means of which a certain group formed by people, things, events, etc., is taken as a sample, which are not considered individually but as part of a whole, which is in turn a proportional representation of the universality of samples available in the field.

Now, since this type of sampling allows to embrace large groups of sample units, data are not always obtained from all the components of the cluster, but from those necessary to be able to quantify the desired statistics.

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Round of 248,682 to the nearest hundred
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248,700

682 is closer to 700 than to 600

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A maker of a certain brand of low-fat cereal barsclaims that the average saturated fat content is 0.5gram. In a random sample of
finlep [7]

Answer:

t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386    

p_v =2*P(t_{(7)}  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance.  So then the claim makes sense

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=0.475

The sample deviation calculated s=0.183

We need to conduct a hypothesis in order to check if the true mean is 0.5 or no, the system of hypothesis would be:  

Null hypothesis:\mu = 0.5  

Alternative hypothesis:\mu \neq 0.5  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=8-1=7  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(7)}  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance.  So then the claim makes sense

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Answer:

Step-by-step explanation:

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I 1   3|

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sukhopar [10]

Answer:

x \geqslant  1

Step-by-step explanation:

The expression in the square root should be greater than or equal to zero, because square root of a negative number does not give a real number.

x  - 1 \geqslant 0 \\ x \geqslant  1

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