In a study comparing banks in Germany and Great Britain, a sample of 145 matched pairs of banks was formed. Each pair contained one bank from Germany and one from Great Britain. The pairings were made in such a way that the two members were as similar as possible in regard to such factors as size and age. The ratio of total loans outstanding to total assets was calculated for each of the banks. For this ratio, the sample mean difference (German – Great Britain) was 0.0518, and the sample standard deviation of the differences was 0.3055. Test, against a two-sided alternative, the null hypothesis that the two population means are equal. 1. Find the p-value of this test.
In a study comparing banks in Germany and Great Britain, a sample of 145 matched pairs of banks was formed. Each pair contained one bank from Germany and one from Great Britain. The pairings were made in such a way that the two members were as similar as possible in regard to such factors as size and age. The ratio of total loans outstanding to total assets was calculated for each of the banks. For this ratio, the sample mean difference (German – Great Britain) was 0.0518, and the sample standard deviation of the differences was 0.3055. Test, against a two-sided alternative, the null hypothesis that the two population means are equal.
1. Find the p-value of this test.
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t = d¯sdn
t = 0.05180.3055145
t = 2.04175
Now lets find p-value
Using Excel
we got , t = 2.04175
df = n - 1 = 145 - 1 = 144
tails = 2 ( since two tailed test )
So, p-value is ,
P-value = TDIST(2.04175, 144, 2) = 0.043001816
Please explain it clealy.
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