Question

Jump to level 1 2 The mean voltage and standard deviation of 5 batteries from each manufacturer were measured. The results are summarized in the following table. Manufacturer Sample mean voltage (millivolts) Sample standard deviatio 3 А 126 5 B 119 4 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: 259 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 voltage of the batteries made by the two manufacturers is different, therefore this is a two tailed test here. Therefore this test is a test for difference in means which is a two tailed t test here.

b) The standard error here is computed as:

$SE =\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{5^2 + 4^2}{5}} = 2.8636$

The test statistic here is computed as:

$t^* = \frac{\bar X_1 - \bar X_2}{SE} = \frac{126 - 119}{2.8636} = 2.445$

Therefore 2.445 is the required test statistic value here.

c) The degrees of freedom here is computed as:
Df = n₁ + n₂ - 2 = 8

Therefore 8 is the degrees of freedom here.

d) For 8 degrees of freedom, the p-value is obtained from t distribution tables here:
p = 2P( t₈ > 2.445) = 2*0.0201 = 0.0402

As the p-value here is 0.0402 < 0.1 which is the level of significance, therefore the test is significant here and we can reject the null hypothesis here. Therefore we have sufficient evidence here that the voltage of the batteries made by the two manufacturers is different