Question: A survey conducted for a random sample of 100 college students on their monthly cell phone expense has yielded the sample mean of $48.5$ and variance of $441$. Suppose that you want to test if the population mean is equal to $45$. What will be the value of the t-test statistics to test this hypothesis?
1. **State the problem:** We want to test the hypothesis that the population mean monthly cell phone expense is $45$ based on a sample of size $n=100$ with sample mean $\bar{x}=48.5$ and sample variance $s^2=441$.
2. **Hypotheses:**
- Null hypothesis $H_0: \mu = 45$
- Alternative hypothesis $H_a: \mu \neq 45$
3. **Formula for t-test statistic:**
$$ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} $$
where $\mu_0$ is the hypothesized population mean, $s$ is the sample standard deviation, and $n$ is the sample size.
4. **Calculate sample standard deviation:**
$$ s = \sqrt{s^2} = \sqrt{441} = 21 $$
5. **Calculate the standard error:**
$$ SE = \frac{s}{\sqrt{n}} = \frac{21}{\sqrt{100}} = \frac{21}{10} = 2.1 $$
6. **Calculate the t-test statistic:**
$$ t = \frac{48.5 - 45}{2.1} = \frac{3.5}{2.1} $$
7. **Simplify the fraction:**
$$ t = \frac{3.5}{2.1} = \frac{\cancel{3.5}}{\cancel{2.1}} \times \frac{10}{10} = \frac{35}{21} = \frac{\cancel{35}}{\cancel{21}} \times \frac{1}{1} = \frac{5}{3} \approx 1.667 $$
**Final answer:** The value of the t-test statistic is approximately $1.667$, which corresponds to option C.