QUESTION IMAGE
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 2 of 4: compute the value of the $t$ - test statistic. round your answer to three decimal places.
Step1: Calculate the pooled - variance
The formula for the pooled - variance $s_p^2$ when the population variances are equal is $s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}$, where $n_1 = 5$, $s_1 = 38$, $n_2=12$, and $s_2 = 27$.
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Step2: Calculate the t - test statistic
The formula for the t - test statistic for two - sample independent t - test with equal variances is $t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$. Since the null hypothesis is $H_0:\mu_1-\mu_2\leq0$ and the alternative hypothesis is $H_1:\mu_1 - \mu_2>0$, and we assume $\mu_1-\mu_2 = 0$ under the null hypothesis, $\bar{x}_1 = 1180$, $\bar{x}_2 = 1120$, $s_p=\sqrt{919.667}\approx30.326$, $n_1 = 5$, and $n_2 = 12$.
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$3.726$