Question

Assume the mean number of taste buds from the general population is 10,000 with a standard deviation of 850. You take a sample of 10 top chefs and find the mean number of taste buds is 10,900. Assume that the number of taste buds in top chefs is a normally distributed variable and assume the standard deviation is the same as for the general population.

(a) What is the point estimate for the mean number of taste buds for all top chefs?

Enter your answer as an integer.

(b) What is the critical value of z (denoted z*) for a 99% confidence interval?

Use the value from the table or, if using software, round to 3 decimal places.

(c) What is the margin of error (E) for the mean number of taste buds for top chefs in a 99% confidence interval?

Round your answer to the nearest whole number.

Construct the 99% confidence interval for the mean number of taste buds for all top chefs.

(d) What is the lower bound of the interval?

Round your answer to the nearest whole number.

(e) What is the upper bound of the interval?

Round your answer to the nearest whole number.

Answer 1

a. Point estimate of mean is

b. As population standard deviation is known we will use z distribution to find CI

z value for 99% CI is 2.58 as P(-2.58<z<2.58)=0.99

c. Margin of Error $E=z*\frac{\sigma }{\sqrt{n}}=2.58*\frac{850}{\sqrt{10}}=693.487\approx 693$

d. Lower bound is $\overline{x}-E=10900 - 693=10207$

e. Upper bound $\overline{x}+E=10900 +693=11593$