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 4 of 4: state the tests conclusion. answer 1 point reject null hypothesis fail to reject null hypothesis

Explanation:

Step1: Calculate degrees of freedom

$df = n_1 + n_2 - 2=5 + 12- 2=15$

Step2: Find critical - value

For a one - tailed test with $\alpha = 0.1$ and $df = 15$, from the t - distribution table, the critical value $t_{\alpha,df}=t_{0.1,15}=1.341$

Step3: Calculate the test statistic

First, calculate the pooled variance $s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}=\frac{(5 - 1)\times38^2+(12 - 1)\times27^2}{5 + 12-2}=\frac{4\times1444+11\times729}{15}=\frac{5776+8019}{15}=\frac{13795}{15}\approx919.67$
The test statistic $t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$, since $\mu_1-\mu_2 = 0$ (null hypothesis), $\bar{x}_1 = 1180$, $\bar{x}_2 = 1120$, $n_1 = 5$, $n_2 = 12$ and $s_p=\sqrt{919.67}\approx30.33$
$t=\frac{(1180 - 1120)-0}{30.33\sqrt{\frac{1}{5}+\frac{1}{12}}}=\frac{60}{30.33\sqrt{\frac{12 + 5}{60}}}=\frac{60}{30.33\sqrt{\frac{17}{60}}}=\frac{60}{30.33\times0.531}\approx3.73$

Step4: Make a conclusion

Since the calculated $t = 3.73>t_{0.1,15}=1.341$, we reject the null hypothesis.

Answer:

Reject Null Hypothesis