Question
In: Statistics and Probability
Assume that the weights of fourteen bags of carrots (in ounces) selected at random from shelves...
Assume that the weights of fourteen bags of carrots (in ounces) selected at random from shelves of Ingles are as follows: 17.5, 18.2, 17.5, 18.9, 18.3, 14.5, 21.1, 20.2, 14.2, 25.2, 19.3, 17.2, 16.8, 19.1
For the data, can we conclude that the average weight of carrots sold at Ingles is different from 19.0 using a hypothesis test? What assumptions were necessary to do this test? Do you have any information on how valid they might be?
Solutions
orchestra answered 2 years agoRelated Solutions
Suppose that the distribution of the weights of bags of carrots from brand A is N(1.2,0.049)...
Suppose that the distribution of the weights of bags of carrots
from brand A is N(1.2,0.049) and the distribution of the weights of
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from two brands is independent. Selecting bags at random find
a) The probability that the sum of a random
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The weights of four randomly and independently selected bags of tomatoes labeled 5.0 pounds were found...
The weights of four randomly and independently selected bags of
tomatoes labeled 5.0 pounds were found to be 5.3, 5.0, 5.6, and 5.4
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bags of tomatoes. (__,__)
(b). Does the interval capture 5.0 pounds? Is there enough
evidence to reject a mean weight of 5.0 pounds? Explain.
The weights of four randomly and independently selected bags of tomatoes labeled 5.0 pounds were found...
The weights of four randomly and independently selected bags of
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Find a? 95% confidence interval for the mean weight of all bags
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The weights of four randomly and independently selected bags of potatoes labeled 20 pounds were found...
The weights of four randomly and independently selected bags of
potatoes labeled 20 pounds were found to be 20.7, 21.4, 20.9,
and 21.2 pounds. Assume Normality. Answer parts (a) and (b)
below.
a. Find a 95% confidence interval for the mean weight of all
bags of potatoes.
b. Does the interval capture 20 pounds? Is there enough
evidence to reject a mean weight of 20 pounds?
A.The interval does not capture 20 pounds, so there is enough
evidence to reject...
the weights of four randomly and independently selected bags of potatoes labeled 20 pounds were found...
the weights of four randomly and independently selected bags
of potatoes labeled 20 pounds were found 21.7,22,20.5, and 21.9.
Assume Normality. Using a two sided alternative hypothesis should
you be able to reject a hypothesis that the population mean is 20
pounds using a significance level of 0.05 why or why not the
confidence interval is reported here I am 95% confidence the
population mean is between 20.4 and 22.6 pounds. B) test the
hypothesis that the population mean is...
A sample of 10 small bags of the same brand of candies was selected. Assume that...
A sample of 10 small bags of the same
brand of candies was selected. Assume that the population
distribution of bag weights is normal. The weight of each bag was
then recorded. The mean weight was 2 ounces
with a sample standard deviation s=0.2
ounces. We would like to calculate an 80% confidence interval for
the average weight of a sample of size 10 .
(3%) standard error =
(3%) The critical t value for an 80%
confidence interval is...
A sample of 10 small bags of the same brand of candies was selected. Assume that...
A sample of 10 small bags of the same
brand of candies was selected. Assume that the population
distribution of bag weights is normal. The weight of each bag was
then recorded. The mean weight was 2 ounces
with a sample standard deviation s=0.2
ounces. We would like to calculate an 80% confidence interval for
the average weight of a sample of size 10 .
(3%) standard error =
(3%) The critical t value for an 80%
confidence interval is...
A sample of 10 small bags of the same brand of candies was selected. Assume that...
A sample of 10 small bags of the same brand of candies was
selected. Assume that the population distribution of bag weights is
normal. The weight of each bag was then recorded. The mean weight
was 3 ounces with a sample standard deviation s=0.5 ounces. We
would like to calculate an 80% confidence interval for the average
weight of a sample of size 10 .
(3%) standard error =
(3%) The critical t value for an 80% confidence interval...
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