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In: Statistics and Probability
A sample of 61 observations yielded a sample standard deviation of 5. If we want to...
A sample of 61 observations yielded a sample standard deviation of 5. If we want to test Ho: σ²=20, the test statistic is:
Solutions
orchestra answered 2 years agoRelated Solutions
a random sample of 100 observations from a population. with standard deviation 60 yielded a sample...
a random sample of 100 observations from a population. with
standard deviation 60 yielded a sample mean of 110. A. test the
null hypothesis that m = 100 and the alternative hypothesis m >
100 using alpha = .05 and interpret the results b. test the null
against the alternative hypothesis that m isn't equal to 110. using
alpha = .05 and interpret the results c. compare the p values of
the two tests you conducted. Explain why the results...
a random sample of 100 observations from a population. with standard deviation 60 yielded a sample...
a random sample of 100 observations from a population. with
standard deviation 60 yielded a sample mean of 110. A. test the
null hypothesis that m = 100 and the alternative hypothesis m >
100 using alpha = .05 and interpret the results b. test the null
against the alternative hypothesis that m isn't equal to 110. using
alpha = .05 and interpret the results c. compare the p values of
the two tests you conducted. Explain why the results...
A random sample of 100 observations from a population with standard deviation 17.83 yielded a sample...
A random sample of 100 observations from a population with
standard deviation 17.83 yielded a sample mean of 93.
1. Given that the null hypothesis is ?=90 and the alternative
hypothesis is ?>90 using ?=.05, find the following: (a) Test
statistic = (b) P - value: (c) The conclusion for this test is: A.
There is insufficient evidence to reject the null hypothesis B.
Reject the null hypothesis C. None of the above
2. Given that the null hypothesis is...
A random sample of 100 observations from a population with standard deviation 23.99 yielded a sample...
A random sample of 100 observations from a population with
standard deviation 23.99 yielded a sample mean of 94.1
1. Given that the null hypothesis is μ=90μ=90 and the
alternative hypothesis is μ>90μ>90 using
α=.05α=.05, find the following:
(a) Test statistic ==
(b) P - value:
(c) The conclusion for this test is:
A. There is insufficient evidence to reject the
null hypothesis
B. Reject the null hypothesis
C. None of the above
2. Given that the null hypothesis is...
A random sample of 100 observations from a population with standard deviation 14.29 yielded a sample...
A random sample of 100 observations from a population with
standard deviation 14.29 yielded a sample mean of 92.7.
1. Given that the null hypothesis is μ=90 and the alternative
hypothesis is μ>90 using α=.05, find the following:
(a) Test statistic =
(b) P - value:
(c) The conclusion for this test is:
A. There is insufficient evidence to reject the null
hypothesis
B. Reject the null hypothesis
C. None of the above
2. Given that the null hypothesis is...
A random sample of 100 observations from a population with standard deviation 70 yielded a sample...
A random sample of 100 observations from a population with
standard deviation 70 yielded a sample mean of 113. Complete parts
a through c below.
a. Test the null hypothesis that muequals100 against the
alternative hypothesis that mugreater than100, using
alphaequals0.05. Interpret the results of the test. What is the
value of the test statistic?
b. test the null hypothesis that mu = 100 against the
alternative hypothesis that mu does not equal 100, using alpha=.05.
interpret the results of...
A random sample of 100 observations from a population with standard deviation 22.99 yielded a sample...
A random sample of 100 observations from a population with
standard deviation 22.99 yielded a sample mean of 94.1. 1. Given
that the null hypothesis is μ≤90 and the alternative hypothesis is
μ>90 using α=.05, find the following: (a) Test statistic = (b) P
- value: (c) The decision for this test is: A. Fail to reject the
null hypothesis B. Reject the null hypothesis C. None of the above
2. Given that the null hypothesis is μ=90 and the...
(1 point) A random sample of 100100 observations from a population with standard deviation 19.788150778587319.7881507785873 yielded...
(1 point) A random sample of 100100 observations from a
population with standard deviation 19.788150778587319.7881507785873
yielded a sample mean of 93.893.8.
(a) Given that the null hypothesis is
?=90μ=90 and the alternative hypothesis is ?>90μ>90 using
?=.05α=.05, find the following:
(i) critical z/t
score
equation editor
Equation Editor
(ii) test statistic ==
(b) Given that the null hypothesis is
?=90μ=90 and the alternative hypothesis is ?≠90μ≠90 using
?=.05α=.05, find the following:
(i) the positive critical z/t
score
(ii) the negative critical z/t
score
(iii) test statistic ==...
(1 point) A random sample of 100100 observations from a population with standard deviation 13.8313.83 yielded...
(1 point) A random sample of 100100 observations from a
population with standard deviation 13.8313.83 yielded a sample mean
of 92.392.3.
1. Given that the null hypothesis is μ=90μ=90 and the
alternative hypothesis is μ>90μ>90 using α=.05α=.05, find the
following:
(a) Test statistic ==
(b) P - value:
(c) The conclusion for this test is:
A. Reject the null hypothesis
B. There is insufficient evidence to reject the
null hypothesis
C. None of the above
2. Given that the null...
The formula for a sample Standard Deviation is Say we want to use standard deviation as...
The formula for a sample Standard
Deviation is
Say we want to use standard
deviation as a way of comparing the amount of spread present in
each of two different distributions.
What is the effect of squaring the deviations? (1
mark)
How does this help us when we compare the spreads of two
distributions? (1 mark)
With reference to the formula and the magnitude of data values,
explain why introducing an outlier to a dataset affects the
Standard Deviations more...
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