Question

A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. The mean braking distance for SUVs equipped with tires made with compound 1 is 52 feet, with a population standard deviation of 7.2. The mean braking distance for SUVs equipped with tires made with compound 2 is 55 feet, with a population standard deviation of 8.8. Suppose that a sample of 70 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ1 be the true mean braking distance corresponding to compound 1 and μ2 be the true mean braking distance corresponding to compound 2. Use the 0.1 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 three decimal places.
Step 4 of 4: Make the decision for the hypothesis test.

Answer 1

We conclude that the mean braking distance for SUV's equipped with tires using compound 1 is shorter than the braking distance when compound 2.

Explanation:

We are given that the mean braking distance for SUV's equipped with tires made with compound 1 is 52 feet, with a population standard deviation of 7.2. The mean braking distance for SUV's equipped with tires made with compound 2 is 55 feet, with a population standard deviation of 8.8.

Suppose that a sample of 70 braking tests are performed for each compound.

Let
`\mu_1` = true mean braking distance corresponding to compound 1.

`\mu_2` = true mean braking distance corresponding to compound 2.

So, Null Hypothesis,
`H_0` :
`\mu_1-\mu_2\geq` = 0 or
`\mu_1\geq\mu_2` {means that the mean braking distance for SUV's equipped with tires using compound 1 is longer or equal to the braking distance when compound 2}

Alternate Hypothesis,
`H_A` :
`\mu_1-\mu_2` < 0 or
`\mu_1<\mu_2` {means that the mean braking distance for SUV's equipped with tires using compound 1 is shorter than the braking distance when compound 2}

The test statistics that would be used here Two-sample z test statistics;

T.S. =
`\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{(\sigma_1^(2) )/(n_1)+(\sigma_2^(2))/(n_2) } }` ~ N(0,1)

where,
`\bar X_1` = sample mean braking distance for SUV's equipped with tires made with compound 1 = 52 feet

`\bar X_2` = sample mean braking distance for SUV's equipped with tires made with compound 2 = 55 feet

`\sigma_1` = population standard deviation of tires made with compound 1 = 7.2

`\sigma_2` = population standard deviation of tires made with compound 2 = 8.8

`n_1` = sample of braking tests for compound 1 = 70

`n_2` = sample of braking tests for compound 2 = 70

So, test statistics =
`\frac{(52-55)-(0)}{\sqrt{(7.2^(2) )/(70)+(8.8^(2))/(70) } }`

= -2.21

The value of z test statistics is -2.21.

Now, at 0.1 significance level the z table gives critical value of -1.282 for left-tailed test.

Since our test statistics is less than the critical value of z as -2.21 < -1.282, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean braking distance for SUV's equipped with tires using compound 1 is shorter than the braking distance when compound 2.