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 12 phones from the manufacturer had a mean range of 1150 feet with a standard deviation of 27 feet. A sample of 7 similar phones from its competitor had a mean range of 1100 feet with a standard deviation of 23 feet. Do the results support the manufacturer's claim? Let μ1μ1 be the true mean range of the manufacturer's cordless telephone and μ2 be the true mean range of the competitor's cordless telephone. Use a significance level of α=0.01 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 H0. Round your answer to three decimal places.

Step 4 of 4: State the test's conclusion.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ1 = μ2
Alternative Hypothesis, Ha: μ1 > μ2

Pooled Variance
sp = sqrt((((n1 - 1)*s1^2 + (n2 - 1)*s2^2)/(n1 + n2 - 2))*(1/n1 + 1/n2))
sp = sqrt((((12 - 1)*27^2 + (7 - 1)*23^2)/(12 + 7 - 2))*(1/12 + 1/7))
sp = 12.2035

Test statistic,
t = (x1bar - x2bar)/sp
t = (1150 - 1100)/12.2035
t = 4.097

Rejection Region
This is right tailed test, for α = 0.01 and df = n1 + n2 - 2 = 17
Critical value of t is 2.567.
Hence reject H0 if t > 2.567

Reject the null hypothesis
There is sufficient evidence to conclude that the calling range (in feet) of its 900-MHz cordless telephone is greater than that of its leading competitor.