An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 79 type K batteries and a sample of 75 type Q batteries. The mean voltage is measured as 9.14 for the type K batteries with a standard deviation of 0.678, and the mean voltage is 9.45 for type Q batteries with a standard deviation of 0.518. Conduct a hypothesis test for the conjecture that the mean voltage for these two types of batteries is different. Let μ1μ1 be the true mean voltage for type K batteries and μ2μ2 be the true mean voltage for type Q batteries. Use a 0.01 level of significance.
Step 2 of 4 : Compute the value of the test statistic. Round your answer to two decimal places.
Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to two decimal places.
Step 4 of 4: Make the decision for the hypothesis test.
And from the data, we know the mean and variance of the sample only.
So we can use a t-test to estimate .
Significance level = 0.01
x1 = 9.14, x2 = 9.45
s1 = 0.678, s2 = 0.518
n1 = 79, n2 = 75
Standard error SE = = 0.097
Degrees of freedom is calculated by Welch-Satterthwaite approximation
df = 164
and finally our t-statistic t = (X1-X2)/SE = (9.14-9.45) / 0.097 = -3.19
For t=-3.19 and df = 164, p value comes out to be 0.0016
Since p<0.05 hence we have strong evidence against the null hypothesis.
So the null hypothesis is rejected and we can claim that the mean voltages of the two batteries are not equal.
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