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Question 9 of 15 Step 1 of 4 01:56:27 An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 66 type K batteries and a sample of 59 type Q batteries. The mean voltage is measured as 8.62 for the type K batteries with a standard deviation of 0.628, and the mean voltage is 9.01 for type Q batteries with a standard deviation of 0.543. Conduct a hypothesis test for the conjecture that the mean voltage for these two types of batteries is different. Let u be the true mean voltage for type K batteries and uz be the true mean voltage for type Q batteries. Use a 0.1 level of significance. Step 1 of 4: State the null and alternative hypotheses for the test. Answer 2 Points 5 Tables Keypad - M2 HOME Ηα: μι Prev Next Activate Windows Go to Settings to activate Windows.

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 H0H0. Round the numerical portion of your answer to three decimal places.

Step 4 of 4:

Make the decision for the hypothesis test.

Answer 1

An engineer is comparing voltages for two types of batteries K and Q using a sample of 66 type K batteries and a sample of 59 type Q batteries.

The mean voltage is measured as 8.62 for the type K batteries with a standard deviation of 0.628.

The mean voltage is 9.01 for type Q batteries with a standard deviation of 0.543

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

Now,

For type k batteries for type Q batteries. Here ă = 8.62 = 9.01 Dis 0.628 02 50.543 ni = 66 n2=59 ry Hypothesis test;- Null hypothesis. Ho ili =12 Hernative hypothesis, HA = Ili Illa © Test -statistic :- 2=2,-ão - 8.62-9.01. 170.62872 (0.543) 66 59 Jorget content fogom (20.990) = -0.39 - 3.4 0.1047 2-[-3.75] ☺ we calculate Degree of freedom. 2 . 2.2 .:DF Foto) t6m1, (0212) ni na n n2-1 2 (o. 0.628 166 = (40.625137 (0.54923 [60.62873/6606760.543%50] 59 66-1 59-1 = 122

critical value with DF 1228 significance Level 0.10 is t 1.289 As the test statistic value is less than the lower critical value i.e-3.75k-1.289 so we reject the null hypothesis.
4) At 0.10 significance level there is sufficient evidence to support the claim that the mean voltage for these two types of batteries is different.