Question

A manufacturer claims that the calling range (in feet) of its 900-MHz cordless telephone is greater than that of its leading competitor. A sample of 1919 phones from the manufacturer had a mean range of 11101110 feet with a standard deviation of 2222 feet. A sample of 1111 similar phones from its competitor had a mean range of 10601060 feet with a standard deviation of 2323 feet. Do the results support the manufacturer's claim? Let μ1μ1 be the true mean range of the manufacturer's cordless telephone and μ2μ2 be the true mean range of the competitor's cordless telephone. Use a significance level of α=0.1α=0.1 for the test. Assume that the population variances are equal and that the two populations are normally distributed.

Step 1 of 4:

State the null and alternative hypotheses for the test.

Step 2 of 4:

Compute the value of the t test statistic. Round your answer to three decimal places.

Step 3 of 4:

Determine the decision rule for rejecting the null hypothesis H0H0. Round your answer to three decimal places.

Step 4 of 4:

State the test's conclusion.

Answer 1

The pooled variance(s_p²) = ((n1 - 1)s1^2 + (n2 - 1)s2^2)/(n1 + n2 - 2)

                                        = (18 * (22)^2 + 10 * (23)^2)/(19 + 11 - 2)

                                        = 500.0714

The test statistic is

   

DF = 19 + 11 - 2 = 28

At = 0.1, the critical value is t_0.1,28 = 1.313

Sine the test statistic value is greater than the critical value(5.902 > 1.313), so we should reject the null hypothesis.

At 0.10 significance level, there is sufficient evidence to support the manufacturer's claim that the calling range of its 900-MHz cordless telephone is greater than of its leading competitor.