Question

4) At Starbucks the advertised nutrition facts say that a Tall Chai latte with nonfat

milk has 32 grams of sugar, which is equivalent to 8 sugar cubes. Of course, there

will be some variability in sugar content. Suppose that the standard deviation is

about 0.1 grams. Engineers monitoring production processes assume that

measurements are normally distributed.

A quality control engineer randomly samples 10 Tall Chai Lattes and finds a

mean sugar content of32.07 grams. ls the difference between 32.07 and 32

statistically significant for a sample of 10?

a) Explain why we can use a normal curve to model the distribution of sample

means despite the fact that the samples only contain 10 lattes.

b) What is the z-score for the engineer's sample? What does the z-score tell us?

c) Is the sample statistically significant? Support your answer using probability.

Answer 1

Answer:

Given that:

At Starbucks the advertised nutrition facts say that a Tall Chai latte with nonfat

milk has 32 grams of sugar, which is equivalent to 8 sugar cubes. Of course, there

will be some variability in sugar content. Suppose that the standard deviation is

about 0.1 grams. Engineers monitoring production processes assume that

measurements are normally distributed.

a) Explain why we can use a normal curve to model the distribution of sample

means despite the fact that the samples only contain 10 lattes.

it is assumed that the measurements are normally distributed. also, the population standard deviation is given as 0.1 grams. Hence it is sufficient to use a one-sample z test.

in other words, the normal curve model is used besides small sample size as the population standard deviation is known.

b) What is the z-score for the engineer's sample? What does the z-score tell us

test statistic, z = (mean-u)/(sd/sqrt(n))
z = (32.07-32)/(0.1/sqrt(10))
z = 2.2136
This tells us that the engineer's sample is 2.2 sd more than the population means.

c) Is the sample statistically significant? Support your answer using probability.

Ho: u = 32
h1: u =/= 32
z = 2.2136

p-value
2*(1-P(z<|z|)
2*(1-P(z<abs(2.2136))
normsdist(abs(2.2136))
0.0269

With z=2.21, p<5%, i reject Null hypothesis and conclude that u =/= 32. there is no evidence to conclude that there are 32 grams of sugar.