Question

You randomly select 16 restaurants and measure the temperature of the coffee sold at each. The sample mean temperature is 162 degrees with a standard deviation of 10 degrees. Assuming that the temperature is approximately normally distributed, a 95% confidence interval for the mean temperature is:________

Answer 1

The 95% confidence interval for the mean coffee temperature is approximately 148°F to 176°F.

Step-by-step explanation:

Identify relevant values:

Sample mean (μ) = 162°F

Standard deviation (σ) = 10°F

Confidence level = 95%

Calculate critical z-score:

For a 95% confidence level, the z-score is 1.96 (from standard normal distribution tables).

Calculate margin of error:

Margin of error (ME) = z-score * standard deviation

ME = 1.96 * 10°F ≈ 19.6°F

Construct the confidence interval:

Lower bound = Sample mean - Margin of error = 162°F - 19.6°F ≈ 142.4°F

Upper bound = Sample mean + Margin of error = 162°F + 19.6°F ≈ 181.6°F

Round to whole degrees: 142°F to 182°F

Therefore, we can be 95% confident that the true mean coffee temperature in all restaurants falls within the range of 148°F to 176°F.

Answer 2

( 157.1 , 166.9 )

Step-by-step explanation:

95% Confidence Interval is the range around sample mean(s), which is 95 certain to contain population parameter (population mean).

Formula : x' + z (s / √n) ,

where x' = sample mean = 162 ; sample standard deviation = 10 ; n = sample size = 16

162 + 1.96 (10 /√ 16)

162 + 4.9

( 157.1 , 166.9 )