A study was conducted to evaluate the stress level of senior business students at a particular college. Forty students were selected at random from the senior business class, and their stress level was monitored by attaching an electrode to the frontalis muscle (forehead). For the forty students, the mean EMG (electromyogram) activity was found to be 35.8 microvolts. In addition, the standard deviation of the EMG readings was found to be 2.5 microvolt. What would be the 99% confidence interval on the true mean EMG activity for all seniors in the class?

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Answer:

The 99% confident that the population mean (μ) falls between 34.73 and 36.86.

Step-by-step explanation:

Consider the provided information.

The formula used for confidence interval on the true mean is:

[tex]\bar{x}\pm t_{n-1}\frac{s}{\sqrt{n}}[/tex]

It is given that mean is 35.8, standard deviation is 2.5, Sample size is 40.

As the sample size is 40 and confidence interval is 99%. Therefore the value of [tex]t_{n-1}=2.70[/tex].

Now, substitute the respective values in the above formula.

[tex]35.8\pm 2.70\times \frac{2.5}{\sqrt{40}}[/tex]

[tex]35.8\pm 2.70\times \frac{2.5}{6.32}[/tex]

[tex]35.8\pm 2.70\times 0.396[/tex]

[tex]35.8\pm 1.068[/tex]

[tex]35.8+1.068[/tex] or [tex]35.8-1.068[/tex]

[tex]36.86[/tex] or [tex]34.73[/tex]

Hence, the 99% confident that the population mean (μ) falls between 34.73 and 36.86.