A population a normally distributed with standard deviation 10. Test the claim t

Question

A population a normally distributed with standard deviation 10. Test the claim that the mean of the population is 68 if a random sample of 121 has mean 69.5. Alpha=0.05. What is the statistical technique use to compare the means of three or more populations. What is required of the populations and the samples used in the technique. State a disadvantage (a weakness) of the mean as a representative of a data set. The mean is a measure of central tendency in a data set. Name two measures of variation in a data set. What is the minimal sample size needed to estimate the mean of a population to within 3 of its true value. The standard deviation of the population is 4. Alpha = 0.05. Determine the maximum error at alpha = 0.05 in estimating the population of a population that favors candidate A if in a sample of 4000, 800 favor candidate A.

Explanation / Answer

4) Given that sd = 10

Population mean (mu) = 68

sample size (n) = 121

Sample mean (Xbar) = 69.5

alpha = 0.05

H0 : mu = 68

H1 : mu not= 68

Critical value we can find by using EXCEL.

syntax is :

=NORMSINV(probability)

probability = alpha

There are two critical values -1.645 and 1.645.

The test statistic is,

Z = (Xbar - mu) / (sd / sqrt(n))

Z = (69.5-68) / (10/sqrt(121)) = 1.650

Z > critical value

Reject H0 at 5% level of significance.

Conclusion : Population mean is differ than 68.

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5) What is the statistical technique use to compare the means of three or more populations.

Here we use one way ANOVA for testing more than two means.

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Disadvantage of mean :

The important disadvantage of mean is that it is sensitive to extreme values/outliers, especially when the sample size is small. Therefore, it is not an appropriate measure of central tendency for skewed distribution.

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Here we are given that sigma = 4

Margin of error (E) = 3

alpha = 0.05

Here we have to calculate sample size n.

n = [ (Zc*sigma) / E ]2

where Zc is critical value for normal distribution.

Zc we can find by using EXCEL.

syntax :

=NORMSINV(probability)

where probability = 1 -a/2

Zc = 1.96

n = [ (1.96*4) / 3 ]2 =  6.8295

Which is approximately 7.

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Here we are given that alpha = 0.05

n = 4000

x = 800

p = x/n = 800 / 4000 = 0.2

Here we have to find E.

We can find E by using formula of n.

n = p*(1-p) * (Zc/E)2

4000 = 0.2*(1-0.2) * (1.96/E)2

4000 = 0.16*(1.96/E)2

4000 / 0.16 = (1.96 / E)2

sqrt(25000) = (1.96 / E)

1.96/E = 158.1139

E = 1.96/158.1139 = 0.0124

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