Two weights, each labeled as weighing 100 g, are each weighed several times on t
ID: 3158158 • Letter: T
Question
Two weights, each labeled as weighing 100 g, are each weighed several times on the same scale. The results, in units of g above 100 g, are as follows:
First weight: 54 88 90 63 40 66
Second weight: 23 39 27 2 49
Since the same scale was used for both weights, and since both weights are similar,
it is reasonable to assume that the variance of the weighing does not depend on the object being weighed. Can you conclude that the weights differ?
Since the Pvalue is-------------------------------- , we----(CAN CANT )------------ conclude that the weights differ.
Explanation / Answer
Formulating the null and alternative hypotheses,
Ho: u1 - u2 = 0
Ha: u1 - u2 =/ 0
At level of significance = 0.05
As we can see, this is a two tailed test.
Calculating the means of each group,
X1 = 66.83333333
X2 = 28
Calculating the standard deviations of each group,
s1 = 19.41562944
s2 = 17.77638883
Thus, the pooled standard deviation is given by
S = sqrt[((n1 - 1)s1^2 + (n2 - 1)(s2^2))/(n1 + n2 - 2)]
As n1 = 6 , n2 = 5
Then
S = 18.70482212
Thus, the standard error of the difference is
Sd = S sqrt (1/n1 + 1/n2) = 11.32633226
As ud = the hypothesized difference between means = 0 , then
t = [X1 - X2 - ud]/Sd = 3.428588571
df = n1 + n2 - 2 = 9
Getting the p value using technology, as this is two tailed,
p = 0.003762508 [ANSWER, P VALUE]
Hence,
Since the Pvalue is [[0.003762508]] which is very small, we [[CAN]] conclude that the weights differ. [CONCLUSION]
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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!
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