Pls answer ASAP. :))) I have no idea what I’m doing
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Answer:
(29.46 mm, 29.54 mm).
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 43 - 1 = 42
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 42 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.018
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 29.5 mm - 0.04mm = 29.46 mm
The upper end of the interval is the sample mean added to M. So it is 29.5 mm + 0.04mm = 29.54 mm
The 95% confidence level for the true mean width is: (29.46 mm, 29.54 mm).
The mass of the lamina is given by
