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Answer:
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.
<u>Sample 1 </u>(type K)
n₁= 37
sample mean x₁[bar]= 8.54
standard deviation S₁= 0.225
<u>Sample 2 </u>(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:<u> (x₁[bar] - x₂[bar]) - (μ₁ - μ₂)</u> ≈ 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!