Question

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 5555 type K batteries and a sample of 7272 type Q batteries. The type K batteries have a mean voltage of 9.119.11, and the population standard deviation is known to be 0.6480.648. The type Q batteries have a mean voltage of 9.439.43, and the population standard deviation is known to be 0.2710.271. 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 2 of 5:

Compute the value of the test statistic. Round your answer to two decimal places.

Answer 1

Step-1:

Let μ1 be the true mean voltage for type K batteries and μ2 be the true mean voltage for type Q batteries.

Hypothesis:

Null Hypothesis Ho: μ1=μ2 ( ie:The mean voltage for these two types of batteries ( K and Q ) is EQUAL.)

Alternative Hypothesis Ha: μ1μ2 (ie: The mean voltage for these two types of batteries ( K and Q ) is DIFFERENT.)

Step-2 :

Test statistic to be used is Z test for two means. The formula as given below;

Step-3:

As we know K batteries are denoted by '1' and Q batteries as "2".

Given values are as follows;

n1= 5555, x-bar1=9.11 , sigma-1=0.648

n2= 7272, x-bar2=9.43 , sigma-2=0.271

Substitute the above given values in to the formula in step-2; we get;

Z = (9.11-9.43)/sqrt((0.648^2/5555)+(0.271^2/7272)) = -34.57

Step-4:
Z critical value at alpha = 0.10 as per following z-table ( two tailed );

We look at the area 0.4500 ( two-tailed means alpha=0.10/2 = 0.05 ==> Rejection area=0.05 and Z=0.5000-0.05 = 0.4500)

Z-critical value = 1.65 & -1.65 ( two tailed test )

Step-5:

Since Z statistic value=-34.57 goes beyond Z critical value ==> Reject Ho

Conclusion :

μ1μ2 (ie: The mean voltage for these two types of batteries ( K and Q ) is DIFFERENT.)

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