An engineer is comparing voltages for two types of batteries (k and q) using a sample of 37 type k batteries and a sample of 58 type q batteries. the mean voltage is measured as 8.54 for the type k batteries with a standard deviation of 0.225, and the mean voltage is 8.69 for type q batteries with a standard deviation of 0.725. 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 0.10.1 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.

Step-by-step explanation:

Hello!

You want to test two samples of batteries to see is the mean voltage of these battery types are different.

Sample 1 (type K)

n₁= 37

sample mean x₁[bar]= 8.54

standard deviation S₁= 0.225

Sample 2 (Type Q)

n₂= 58

sample mean x₂[bar]= 8.69

standard deviation S₂= 0.725

1. The test hypothesis are:

H₀: μ₁ = μ₂

H₁: μ₁ ≠ μ₂

2. I'll apply the Central Limit Theorem and approximate the distribution of the sample means to normal so that I can use an approximate Z statistic for this test.

Z: (x₁[bar] - x₂[bar]) - (μ₁ - μ₂) ≈ N(0;1)

        √ (S₁²/n₁) + (S₂²/n₂)

= (8.54 - 8.69) / [√ (0.225²/37) + (0.725²/58)]

= -1.468 ≅ -1.47

3. This is a two tailed test, so you'll have two critical values

You'll reject the null hypothesis if ≤ -1.96 or if ≥ 1.96

You'll not reject the null hypothesis if -1.96 < < 1.96

4.

Since the value = -1.47 is in the acceptance region, the decision is to not reject the null hypothesis.

I hope it helps!

mean, 40.6

step-by-step explanation: