Question

one cable company claims that it has excellent customer service. in fact, the company advertises that a technician will arrive within 45 minutes after a service call is placed. one frustrated customer believes this is not accurate, claiming that it takes over 45 minutes for the cable technician to arrive. the customer asks a simple random sample of 25 other cable customers how long it has taken for the cable technician to arrive when they have called for one. the sample mean for this group is 47.9 minutes with a standard deviation of 7.6 minutes. assume that the population distribution is approximately normal. test the customers claim at the 0.01 level of significance. step 2 of 3: compute the value of the test statistic. round your answer to three decimal places.

Explanation:

Step1: Identify the formula

For a one - sample t - test, the test statistic formula is $t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean under the null hypothesis, $s$ is the sample standard deviation, and $n$ is the sample size.

Step2: Define the values

We are given that $\bar{x} = 47.9$, $\mu=45$, $s = 7.6$, and $n = 25$.

Step3: Calculate the test statistic

Substitute the values into the formula:
\[

$$\begin{align*} t&=\frac{47.9 - 45}{\frac{7.6}{\sqrt{25}}}\\ &=\frac{2.9}{\frac{7.6}{5}}\\ &=\frac{2.9}{1.52}\\ &\approx1.908 \end{align*}$$

\]

Answer:

$1.908$