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GuDViN [60]
3 years ago
11

A 90% confidence interval for the mean height of students

Mathematics
1 answer:
Westkost [7]3 years ago
4 0

Answer:

ME= \frac{69.397-60.128}{2}= 4.6345 \approx 4.635

And the best answer on this case would be:

b) m = 4.635

Step-by-step explanation:

Let X the random variable of interest and we know that the confidence interval for the population mean \mu is given by this formula:

\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}

The confidence level on this case is 0.9 and the significance \alpha=1-0.9=0.1

The confidence interval calculated on this case is 60.128 \leq \mu \leq 69.397

The margin of error for this confidence interval is given by:

ME =t_{\alpha/2} \frac{s}{\sqrt{n}}

Since the confidence interval is symmetrical we can estimate the margin of error with the following formula:

ME = \frac{Upper -Lower}{2}

Where Upper and Lower represent the bounds for the confidence interval calculated and replacing we got:

ME= \frac{69.397-60.128}{2}= 4.6345 \approx 4.635

And the best answer on this case would be:

b) m = 4.635

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