Homework issue

I’m going through the homework that is due for Tuesday and i’m not getting a correct answer. Can anyone see what the issue is? Maybe its my format or something.

Here was the question

You measure 42 watermelons’ weights, and find they have a mean weight of 56 ounces. Assume the population standard deviation is 12.9 ounces. Based on this, construct a 95% confidence interval for the true population mean watermelon weight

My answers (which I feel was correct and I also did this in R-Studio to double check.)

s = 12.9
se = s/sqrt(42)
se
# Out:
# [1] 1.990513

me = 2*se
me
# Out:
[1] 3.981026

My math was as followed.

Standard deviation = 12.9
Mean = 56
Sample Size = 42
Confidence Interval = 95%

SE = 12.9 / sqrt(42)
SE = 1.990513

ME = 2 * 1.990513
ME = 3.98

2 in the ME because of the confidence interval with the rule of thumb of probability is 95%.

So the answer SHOULD be 56 ± 3.98

What do you think?

@jthomps6 I think your work looks great, given the presentation we had in class the other day. But, the choice of multiplier z^*=2 that we learned was based on the rules of thumb for the normal distribution namely:

• 68\% of the population lies within 1 standard deviation of the mean,
• 95\% of the population lies within 2 standard deviations of the mean, and
• 99.7\% of the population lies within 3 standard deviations of the mean.

I guess that MyOpenMath is expecting us to use a more precise approach by examining the normal tables to find z^*. We’ll discuss this in class tomorrow but, the short story for now is:

• Try \mathbb z^* = 1.96