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 10 phones from the manufacturer had a mean range of 1020 feet with a standard deviation of 25 feet. A sample of 17 similar phones from its competitor had a mean range of 1010 feet with a standard deviation of 29 feet. Do the results support the manufacturer's claim? Let μ1 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.a. State the null and alternative hypotheses for the test.b. Compute the value of the t test statistic. Round your answer to three decimal places.c. Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places.d. State the test's conclusion.

Answer 1

(a) Null hypothesis: The calling range (in feet) of a manufacturer's 900-MHz cordless telephone is the same as that of its leading competitor

Alternate hypothesis: The calling range (in feet) of a manufacturer's 900-MHz cordless telephone is greater than that of its leading competitor

(b) The t-test statistic is 1.490

(c) The decision rule is 2.787

(d) The calling range (in feet) of a manufacturer's 900-MHz cordless telephone is greater than that of its leading competitor

Explanation:

(a) Null hypothesis is a statement from a population parameter that is subject to testing while alternate hypothesis is a statement that negates the null hypothesis

(b) Mean of manufacturer = 1020, n1 = 10, sd = 25, variance = 25^2 = 625

Mean of competitor = 1010, n2 = 17, sd = 29, variance = 29^2 = 841

Pooled variance = 625(10-1) + 841(17-1) ÷ (10+17-2) = (625×9) + (841×16) ÷ (27-2) = (5625+13456) ÷ 25 = 19081 ÷ 25 = 763.24

t = (mean of manufacturer - mean of competitor) ÷ √[pooled variance (1/n1 + 1/n2)] = (1020 - 1010) ÷ √[763.24(1/10 + 1/17)] = 10 ÷ √45.031 = 10 ÷ 6.711 = 1.490

(c) Total number of observation = n1 + n2 = 10+17 = 27, degree of freedom = 27 - 2 = 25

t-value corresponding to 25 degrees of freedom and 0.01 significance level is 2.787

(d) The t-test statistic (1.490) is less than the critical value (2.787), so reject the null hypothesis

Conclusion: The calling range (in feet) of a manufacturer's 900-MHz cordless telephone is greater than that of its leading competitor