Question

A physicist is testing a new device that produces photon waves. The wavelengths of photons produced with this method are known to be normally distributed with standard deviation 2.7 microns. If he/she measures 15 waves that were produced and gets a sample mean 49 microns, what upper bound would be reported for the 97% confidence interval for the mean wavelength in microns? Round your answer to 4 decimal places.

Answer 1

To find the upper bound of the 97% confidence interval for the mean wavelength, use the formula 'upper bound = sample mean + z * (standard deviation / sqrt(sample size))'. Substituting the given values, the upper bound would be approximately 50.8811 microns.

Step-by-step explanation:

To find the upper bound of the 97% confidence interval for the mean wavelength, we can use the formula:

Upper bound = sample mean + z * (standard deviation / sqrt(sample size))

In this case, the sample mean is 49 microns, the standard deviation is 2.7 microns, and the sample size is 15. The critical value z for a 97% confidence interval is approximately 1.8808.

Substituting these values into the formula, we get:

Upper bound = 49 + 1.8808 * (2.7 / sqrt(15))

Calculating this, the upper bound would be approximately 50.8811 microns, rounded to 4 decimal places.