SOLUTION: Given the sample mean = 22.625, sample standard deviation = 4.9339, and N = 40 for the low income group,
Test the claim that the mean nickel diameter drawn by children in the low
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-> SOLUTION: Given the sample mean = 22.625, sample standard deviation = 4.9339, and N = 40 for the low income group,
Test the claim that the mean nickel diameter drawn by children in the low
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Question 1183057: Given the sample mean = 22.625, sample standard deviation = 4.9339, and N = 40 for the low income group,
Test the claim that the mean nickel diameter drawn by children in the low income group is greater than 21.21 mm.
Teat at the 0.05 significance level.
a) Identify the correct alternative hypothesis: = > 21.21
Give all answers correct to 3 decimal places.
b) The test statistic value is:
c) Using the Traditional method, the critical value is:
You can put this solution on YOUR website! the sample mean is 22.625 and the sample standard deviation is 4.9339.
the sample size is 40.
test mean is 21.21 and test is whether the sample mean is greater than that.
test is whether sample mean is greater than test mean.
standard error = sample standard deviation / sqrt(sample size) = 4.9339 / sqrt(40) = .780118 rounded to 6 decimal places.
t-score = (x - m) / s
x is the samplemean.
m is the test mean
s is the stnadard error
t = (22.625 - 21.21) / .780118 = 1.814 rounded to 3 decimal places.
critical p-value is .05
test is one tailed, therefore critical p-value is .05 on the high end, because the test is whether the sample mean is greater than the test mean.
h0 = 21.21
ha = 22.625
degrees of freedom = 40 minus 1 = 39
p-value for area to the right of t-value of 1.814 with 39 degrees of freedom is .0387 rounded to 4 decimal places.
since this is less than .05, results are significant and alternate hypothesis is accepted.
