Question

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 75 type K batteries and a sample of 46 type Q batteries. The type K batteries have a mean voltage of 8.51, and the population standard deviation is known to be 0.344. The type Q batteries have a mean voltage of 8.87, and the population standard deviation is known to be 0.644. 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 3 of 4 : Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to two decimal places.

Answer 1

The decision rule for rejecting the null hypothesis, considering the t-distribution, is of:

• |t| < 1.9801 -> do not reject the null hypothesis.

• |t| > 1.9801 -> reject the null hypothesis.

What are the hypothesis tested?

At the null hypothesis, it is tested if there is not enough evidence to conclude that the mean voltage for these two types of batteries is different, that is, the subtraction of the sample means is of zero, hence:

$H_0: \mu_1 - \mu_2 = 0$

At the alternative hypothesis, it is tested if there is enough evidence to conclude that the mean voltage for these two types of batteries is different, that is, the subtraction of the sample means different of zero, hence:

$H_1: \mu_1 - \mu_2 \neq 0$

We have a two-tailed test, as we are testing if the mean is different of a value.

Considering the significance level of 0.05, with 75 + 46 - 2 = 119 df, the critical value for the test is given as follows:

|t| = 1.9801.

Hence the decision rule is:

• |t| < 1.9801 -> do not reject the null hypothesis.

• |t| > 1.9801 -> reject the null hypothesis.

More can be learned about the t-distribution in the test of an hypothesis at brainly.com/question/13873630

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