Question

< Jump to level 1 The mean voltage and standard deviation of 7 batteries from each manufacturer were measured. The results are summarized in the following table. 2 3 Manufacturer Sample mean voltage (millivolts) Sample standard deviatio A 161 3 B 157 2 What type of hypothesis test should be performed? Select What is the test statistic? Ex: 0.123 0 What is the number of degrees of freedom? Ex 250 Does sufficient evidence exist to support the claim that the voltage of the batteries made by the two manufacturers is different at the a = 0.1 significance level? Select 2 Check Next

Answer 1

a) As we are testing here whether the mean voltage of the batteries made by the 2 manufacturers is different, therefore this is a test for difference in two means here.

b) The standard error here is computed as:

$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{3^2 + 2^2}{7}} = 1.3628$

The test statistic now is computed here as:

$t^* = \frac{\bar X_1 - \bar X_2}{SE} = \frac{161 - 157}{1.3628} = 2.935$

Therefore 2.935 is the required test statistic value here.

c) The degrees of freedom here is computed as:
Df = n₁ + n₂ - 2 = 12

Therefore 12 is the degrees of freedom here.

d) For 12 degrees of freedom and 0.1 level of significance, we have from t distribution tables here:
P( t₁₂ < 1.782) = 0.95

Therefore, P(-1.782 < t₁₂ < 1.782) = 0.9

As the test statistic value 2.935 > 1.782, therefore it lies in the rejection region and therefore we reject the null hypothesis here and conclude that we have sufficient evidence here that the voltage of the batteries made by the two manufacturers is different