Question

Firebrick Tire Company wants to increase sales of its long running, low end Roadmaster Tire brand with a new advertising campaign claiming they will last at least 28,000 miles [u=28,000 miles]. Tests with a random sample [n=30 tires] show a sample mean [x=$27,500 miles] with a sample standard deviation [s=1000 miles]. At a .05 level of significance, these tests indicate

a. Reject Ho: Z of -2.7386 less than -1.6450
b. Reject Ho: Z of -2.7386 less than 1.6450
c. Do not reject Ho: Z of 1.6450 less than 2.7386
d. Do not reject Ho: Z of -1.6450 less than 2.7386
e. None of the Above

Answer 1

Answer on Question # 41902, Math, Statistics and Probability

Firebrick Tire Company wants to increase sales of its long running, low end Roadmaster Tire brand with a new advertising campaign claiming they will last at least 28,000 miles [u=28,000 miles]. Tests with a random sample [n=30 tires] show a sample mean [x=$27,500 miles] with a sample standard deviation [s=1000 miles]. At a .05 level of significance, these tests indicate

a. Reject Ho: Z of -2.7386 less than -1.6450

b. Reject Ho: Z of -2.7386 less than 1.6450

c. Do not reject Ho: Z of 1.6450 less than 2.7386

d. Do not reject Ho: Z of -1.6450 less than 2.7386

e. None of the Above

Solution

Step 1. State $H_0$: $\mu \geq 28000$, $H_1$: $\mu < 150$.

Step 2. Type of test - left-tailed test.

Step 3. Level of significance: $\alpha = 0.05$.

Step 4. Critical value of the statistic: $z = -1.6450$.

Step 5. Diagram

Step 6. Decision rule: Reject $H_0$ if t computed from evidence less than -1.6450.

Step 7. Compute the statistic:

Evidence: $n = 30$, $\bar{x} = 27500$, $s = 1000$.

$$z = \frac {\bar {x} - \mu}{\frac {s}{\sqrt {n}}} = \frac {27500 - 28000}{\frac {1000}{\sqrt {30}}} = -2.7386.$$

Step 8. Conclusion:

a. Reject Ho: Z of -2.7386 less than -1.6450.

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