An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 7575 type K batteries and a sample of 4545 type Q batteries. The type K batteries have a mean voltage of 9.279.27, and the population standard deviation is known to be 0.7400.740. The type Q batteries have a mean voltage of 9.619.61, and the population standard deviation is known to be 0.8440.844. 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 0.10.1 level of significance.
Step 1 of 4: State the null and alternative hypotheses for the test.
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.
1)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ1 = μ2
Alternative Hypothesis, Ha: μ1 ≠ μ2
2)
Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(0.5476/75 + 0.712336/45)
sp = 0.1521
Test statistic,
z = (x1bar - x2bar)/sp
z = (9.27 - 9.61)/0.1521
z = -2.24
3)
Rejection Region
This is two tailed test, for α = 0.1
Critical value of z are -1.64 and 1.64.
Hence reject H0 if z < -1.64 or z > 1.64
4)
fail to reject null hypothesis.
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