An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 36 type K batteries and a sample of 51 type Q batteries. The mean voltage is measured as 9.29 for the type K batteries with a standard deviation of 0.486 and the mean voltage is 9.59 for type Q batteries with a standard deviation of 0.430. Conduct a hypothesis test for the conjecture that the mean voltage for these two types of batteries is different. Let μ1 be the true mean voltage for type K batteries and μ2 be the true mean voltage for type Q batteries. Use a 0.05 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. (Reject or Fail to Reject Null Hypothesis)
For type K batteries:
x̅1 = 9.29, s1 = 0.486, n1 = 36
For type Q batteries:
x̅2 = 9.59, s2 = 0.430, n2 = 51
α = 0.05
1. Null and Alternative hypothesis:
Ho : µ1 = µ2
H1 : µ1 ≠ µ2
Pooled variance :
S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((36-1)*0.486² + (51-1)*0.43²) / (36+51-2) = 0.2060
2. Test statistic:
t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (9.29 - 9.59) / √(0.206*(1/36 + 1/51)) = -3.0363
df = n1+n2-2 = 85
3. Critical value :
Two tailed critical value, t crit = T.INV.2T(0.05, 85) = 1.99
Reject Ho if t > 1.99
4. Conclusion:
As |t| = 3.0363 > 1.99, Reject the null hypothesis.
There is enough evidence to conclude that the mean voltage for these two types of batteries is different.
Get Answers For Free
Most questions answered within 1 hours.