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 5 phones from the manufacturer had a mean range of 1180 feet with a standard deviation of 38 feet. a sample of 12 similar phones from its competitor had a mean range of 1120 feet with a standard deviation of 27 feet. do the results support the manufacturers claim? let $mu_1$ be the true mean range of the manufacturers cordless telephone and $mu_2$ be the true mean range of the competitors cordless telephone. use a significance level of $alpha = 0.1$ for the test. assume that the population variances are equal and that the two populations are normally distributed. step 3 of 4: determine the decision rule for rejecting the null hypothesis $h_0$. round your answer to three decimal places. answer 1 point reject $h_0$ if

Explanation:

Step1: Calculate degrees of freedom

The degrees of freedom $df=n_1 + n_2-2$, where $n_1 = 5$ and $n_2=12$. So $df=5 + 12-2=15$.

Step2: Determine the critical - value

Since this is a one - tailed test with $\alpha = 0.1$ and $df = 15$, looking up in the t - distribution table, the critical value $t_{\alpha,df}=t_{0.1,15}=1.341$.

Step3: State the decision rule

The null hypothesis $H_0:\mu_1\leq\mu_2$ and the alternative hypothesis $H_1:\mu_1>\mu_2$. We reject $H_0$ if $t>1.341$.

Answer:

Reject $H_0$ if $t>1.341$