Question

Assume the mean number of taste buds from the general population is 10,000 with a standard deviation of 900. You take a sample of 10 top chefs and find the mean number of taste buds is 11,200. 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 99% 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 99% confidence interval? Round your answer to the nearest whole number.
E =  taste buds

(d) Construct the 99% confidence interval for the mean number of taste buds for all top chefs. Round your answers to the nearest whole number.
< μ <   

(e) Based on your answer to part (d), are you 99% confident that top chefs have, on average, more taste buds than the general population and why?

Yes, because the population mean of 10,000 is below the upper limit of the confidence interval for the mean for top chefs.Yes, because the general population mean of 10,000 is below the lower limit of the confidence interval for the mean for top chefs.     

No, because the general population mean of 10,000 is below the lower limit of the confidence interval for the mean for top chefs.No, because the population mean of 10,000 is below the upper limit of the confidence interval for the mean for top chefs.

(f) Why were we able to use the methods of this chapter despite such a small sample?

Because we are assuming the number of taste buds in top chefs is a normally distributed variable.

Because σ is greater than 100.    

Because the sample mean is sufficiently large.

Because the number of taste buds represents a discrete variable.

Answer 1

a_)

point estimate for the mean number of taste buds for all top chefs =11200

b)

critical value of z =2.58

c)

sample size    n=	10.00
std deviation σ=	900.00
std error ='σx=σ/√n=	284.6050

for 99 % CI value of z=		2.580
margin of error E=z*std error =		734.281 ~ 734

d)

lower bound=sample mean-E=		10465.72
Upper bound=sample mean+E=		11934.28
from above 99% confidence interval for population mean =(10466 <μ< 11934)

e)

.Yes, because the general population mean of 10,000 is below the lower limit of the confidence interval for the mean for top chefs.     

f)Because we are assuming the number of taste buds in top chefs is a normally distributed variable.