To calculate the test statistic, we typically use a t-test formula when dealing with a sample. The formula for the t-test statistic is:
\[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \]
Where:
- \(\bar{x}\) is the sample mean (3.18),
- \(\mu_0\) is the population mean under the null hypothesis (not specified here, so I’ll assume a hypothetical value),
- \(s\) is the sample standard deviation (0.06),
- \(n\) is the sample size (40).
Since the population mean (\(\mu_0\)) is not specified, let's assume \(\mu_0 = 3.0\) for the sake of this exercise.
Substituting the values into the formula gives:
\[ t = \frac{3.18 - 3.0}{\frac{0.06}{\sqrt{40}}} \]
Calculating the values:
- Calculate the standard error (SE):
\[ SE = \frac{0.06}{\sqrt{40}} \approx \frac{0.06}{6.32} \approx 0.00946 \]
- Calculate the test statistic:
\[ t = \frac{3.18 - 3.0}{0.00946} \approx \frac{0.18}{0.00946} \approx 19.0 \]
Test Statistic: 19.0 (to 2 decimals, this would be recorded as 19.00)
Next, we will determine the p-value for our t-test statistic. For a t-distribution with \(n-1 = 39\) degrees of freedom and a test statistic of 19.0, you will typically refer to t-distribution tables or a calculator for the p-value.
Given that a t-value of 19.0 typically corresponds to a very high level of significance, the p-value will be extremely small (close to 0).
P-value: Close to 0 (let's say < 0.01 for practical purposes)
Finally, based on the context of the hypothesis test:
- If the p-value is less than the significance level (commonly set at 0.05), we reject the null hypothesis.
- If the p-value is greater than the significance level, we fail to reject the null hypothesis.
Assuming we are using a significance level of 0.05:
Since the p-value is close to 0, we would:
Reject the null hypothesis.