Question

Taste Buds – Top Chefs: Assume the mean number of taste buds from the general population is 10,000 with a standard deviation of 750. 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?
  taste buds

(b) What is the critical value of z (denoted z_α/2) for a 95% confidence interval? Use the value from the table or, if using software, round to 2 decimal places.
z_α/2 =  

(c) What is the margin of error (E) for the mean number of taste buds for top chefs in a 95% confidence interval? Round your answer to the nearest whole number.
E =   taste buds

(d) Construct the 95% confidence interval for the mean number of taste buds for all top chefs. Round your answers to the nearest whole number.
  < μ <  

Answer 1

Solution :

Given that,

a)

Point estimate = sample mean = = 10900

Population standard deviation = = 750

Sample size = n = 10

b)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

c)

Margin of error = E = Z/2* ( /n)

= 1.96 * (750 / 10 )

= 464.85

= 465

At 95% confidence interval estimate of the population mean is,

- E < < + E

10900 - 465 < < 10900 + 465

10435 < < 11365