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Let's first define the null hypothesis (\(H_{0}\)) and alternative hypothesis (\(H_{1}\)). - \(H_{0}: \mu = 70\) (There is no increase in yield for the new variety) - \(H_{1}: \mu > 70\) (There is an increase in yield for the new variety)

Next, we need to calculate the standard error (SE) of the sample mean. The standard error formula is: $$ SE = \frac{s}{\sqrt{n}} $$ where \(s\) is the sample standard deviation, and \(n\) is the sample size. We are given \(s=12.6\) and \(n=40\). So, $$ SE = \frac{12.6}{\sqrt{40}} \approx 1.99 $$ We also need to calculate the degrees of freedom (df) for the t-test. For a one-sample t-test, the degrees of freedom is \(df = n - 1\). In this case, df is: $$ df = 40 - 1 = 39 $$

Now let's calculate the t-test statistic. The formula for the test statistic is: $$ t = \frac{\bar{x} - \mu_{0}}{SE} $$ where \(\bar{x}\) is the sample mean, \(\mu_{0}\) is the population mean under null hypothesis and SE is the standard error. We have \(\bar{x}=77\), \(\mu_{0}=70\), and \(SE \approx 1.99\). So, the test statistic is: $$ t = \frac{77 - 70}{1.99} \approx 3.52 $$ This is the value of the test statistic for testing the hypothesis that the new variety will increase yield. As for whether the test statistic is likely assuming \(H_{0}\) is true, it is generally considered unlikely if the test statistic falls within the critical region (i.e., the region beyond the critical value). Since we were performing a one-tailed test with the alternative hypothesis stating \(\mu > 70\), we'd compare our test statistic to the critical value corresponding to a given significance level (e.g., 0.05, which is commonly used). In this case, the critical value for a significance level of 0.05 and 39 degrees of freedom can be found in a t-distribution table and is approximately 1.685. Our calculated test statistic, 3.52, is clearly larger than the critical value, which implies that the test statistic is unlikely assuming \(H_{0}\) is true. In conclusion, the test statistic is approximately 3.52, and given that it falls in the critical region, it is unlikely assuming the null hypothesis is true.