Question

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 96 type K batteries and a sample of 98 type Q batteries. The mean voltage is measured as 8.79 for the type K batteries with a standard deviation of 0.661, and the mean voltage is 9.05 for type Q batteries with a standard deviation of 0.206. 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 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. Fail or reject to fail.

Answer 1

This is the two tailed test .

The null and alternative hypothesis is

H₀ :   $\mu$ ₁ = $\mu$ ₂

H_a : $\mu$ ₁$\neq$$\mu$₂

Test statistic = z

= $\bar x$ ₁ - $\bar x$ ₂ / $\sqrt$ [$\sigma^2$₁ / n₁ + $\sigma^2$ ₂ / n₂]

= 8.79 - 9.05 / $\sqrt$ [0.661² / 96 + 0.206²/ 98]

Test statistic = -3.68

Reject H0 |Z| > 1.96

Reject the null hypothesis .