Recently, Blue Diamond claimed that there was an average of at least 130 almonds in their $16 \mathrm{oz}$. bags of almonds. The population standard deviation is known to be 7.5 almonds. You would like to test this claim. You randomly select 36 of their $16 \mathrm{oz}$. bags and count the nuts in each. You find an average of 128 almonds per bag. At the $a=0.05$ level of significance, do you have sufficient evidence to reject their claim?

Step 1 ：This problem is about hypothesis testing for the population mean when the population standard deviation is known. The claim is that the average number of almonds in a $16 \mathrm{oz}$ bag is at least 130. We want to test this claim.

Step 2 ：The null hypothesis is that the population mean is at least 130, and the alternative hypothesis is that the population mean is less than 130.

Step 3 ：We use a one-sample z-test to test the hypotheses. The test statistic is calculated as follows: \[Z = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}}\] where \(\bar{X}\) is the sample mean, \(\mu_0\) is the hypothesized population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

Step 4 ：The critical value for the test is the z-value such that the area to its left under the standard normal curve is \(a\), the level of significance. If the test statistic is less than the critical value, we reject the null hypothesis.

Step 5 ：Given that \(\mu_0 = 130\), \(\sigma = 7.5\), \(n = 36\), \(\bar{X} = 128\), and \(\alpha = 0.05\), we calculate the test statistic \(Z\) to be -1.6 and the critical value to be approximately -1.645.

Step 6 ：Since the test statistic is greater than the critical value, we do not reject the null hypothesis.

Step 7 ：Final Answer: We do not have sufficient evidence at the \(a=0.05\) level of significance to reject the claim that there is an average of at least 130 almonds in their $16 \mathrm{oz}$ bags of almonds. Therefore, the answer is \(\boxed{\text{No}}\).