Thank you for the opportunity to help you with your question!
In stats, whenever you are asked to compare a population mean to a sample mean the first thing that should come to your mind is: t test!
Recall that the t test equation is: (Sample mean - Population mean)/ SEM
In your question above 30g is your "population" mean, it's what is supposed to be the norm, and 29.2 is your sample mean.
since you are not given SEM directly in the question you need to calculate it from the standard deviation.
Recall that the SEM equation is SQ/ sqrt(N). N is the sample number, in your case it is 16. Therefore, the SEM = 4.2/ sqrt(16) which equals 1.05
We can now plug in the SEM to our t test equation, which would look like the following: t= (29.2 - 30)/1.05
the t that we end up getting after you calculate it out is -.7619 (you can ignore the negative sign though)
In order to see if there is a significant difference between the weights of the balls, we use that t score that we calculated and compare it to the t table of critcal values.
We use degrees of freedom in N-1, which in our case is 16-1, which =s 15. We go down the column to df of 15, and across to the .05 level and compare the t value that we found to the critical t value on the table. Our number of .7619 is less than the critical value at df=15, and therefore the difference is not significant and we can refute the claim that the sample of balls comes from a population with the mean weight of less than 30g.
I hope I was able to clarify that for you.
If you have any further questions or need any area explained further don't hesitate to contact me!
I'm always happy to answer your questions!
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