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 :-   $\mu_{1} \leq \mu_{2}$

H1 :-   $\mu_{1} >\mu_{2}$

Test Statistic :-
$t = (\bar{X_{1}} - \bar{X_{2}}) / S_{p}\sqrt{ ( 1 / n1) + (1 / n2)}$

$S_{p} = ( \sqrt{( n1 - 1 )S_{1}^{2} + (n2 - 1 )S_{2}^{2} ) / ( n1 + n2 - 2 )}$
$S_{p} = (\sqrt{( 18 - 1 ) 37^{2} + ( 13 - 1 ) 39 ^{2} ) / ( 18 + 13 - 2 )}$

$t = ( 1230 - 1190) / 37.8404 \sqrt{ ( 1 / 18) + (1 / 13 )}$
t = 2.9042

Test Criteria :-
Reject null hypothesis if $t > t_{\alpha, n1 + n2 - 2}$
$t_{\alpha, n1 + n1 - 2} = t_{ 0.1 , 18 + 13 - 2} = 1.311$
$t > t_{\alpha, n1 + n2 - 2} = 2.9042 > 1.311$
Result :- Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 2.9042 ) = 0.0035
Reject null hypothesis if P value < $\alpha = 0.10$ 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.