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 18 phones from the manufacturer had a mean range of 1230 feet with a standard deviation of 37 feet. A sample of 13 similar phones from its competitor had a mean range of 1190 feet with a standard deviation of 39 feet. Do the results support the manufacturer's claim? Let μ 1 be the true mean range of the manufacturer's cordless telephone and μ 2 be the true mean range of the competitor's cordless telephone. Use a significance level of α=0.1 for the test. Assume that the population variances are equal and that the two populations are normally distributed.

Step 2 of 4 : Compute the value of the t test statistic. Round your answer to three decimal places.

PLEASE SHOW WORK!!!

Answer 1

To Test :-

H0 :-  

H1 :-  

Test Statistic :-

t = 2.9042

Test Criteria :-
Reject null hypothesis if


Result :- Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 2.9042 ) = 0.0035
Reject null hypothesis if P value < level of significance
P - value = 0.0035 < 0.1 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is sufficient evidence to support the claims that the calling range (in feet) of its 900-MHz cordless telephone is greater than that of its leading competitor at 10% level of significance.