Question

A coffee company sells bags of coffee beans with an advertised weight of 454 grams. A random sample of 20 bags of coffee beans has an average weight of 457 grams. Weights of coffee beans per bag are known to follow a normal distribution with standard deviation 5 grams.

(a) Construct a 95% confidence interval for the true mean weight of all bags of coffee beans. (Instead of typing ±, simply type +-.) (1 mark)

(b) Provide an interpretation of the confidence interval in (a). (1 mark)

(c) Conduct a hypothesis test at the 5% level of significance to determine if there is evidence that the true mean weight of all bags of coffee beans differs from 454 grams. You will be marked on your calculation of the appropriate test statistic, P-value and a carefully worded conclusion.

(d) Provide an interpretation of the P-value in (c). (1 mark)

(e) Could the confidence interval in (a) have been used to conduct the test in (c)? Explain why or why not. If it could have been used, what would your conclusion be and why? (1 mark)

Answer 1

n = 20

Sample mean =

Population standard deviation =

Here Population standard deviation is known so we use z test.

a) 95% confidence interval for the true mean weight of all bags of coffee beans is

where zc is z critical value for (1+c)/2 = (1+0.95)/2 = 0.975

zc = 1.96 (From statistical table of z values)

(Round to 2 decimal)

95% confidence interval for the true mean weight of all bags of coffee beans is (454.81, 459.19)

b)

Interpretation of confidence interval:

We are 95% confident that true mean weight of all bags of coffee beans will lie between the interval (454.81, 459.19)

c)

Here we have to test that

Null hypothesis:

Alternative hypothesis:

where

Test statistic:

z = 2.68 (Round to 2 decimal)

Test statistic = z = 2.68

alpha = Level of significance = 0.05

Test is two tailed test.

P value = 2* P(z > 2.68)

= 2 * (1 - P(z < 2.68))

= 2 * ( 1 - 0.9963) (From statistcal table of z values)

= 2 * 0.0037

= 0.0074

P value = 0.0074

Here p value < alpha

So we reject null hypothesis H0.

Conclusion: There is sufficient evidence to conclude that the true mean weight of all bags of coffee beans differs from 454 grams.

d)

P value = 0.0074 = 0.74%

Interpretation of p value:

If the true mean weight of all bags of coffee beans is 454 grams and if we sample the same number of bags repeatedly then 0.74% of samples will result in a sample mean farther from 454 than the sample mean of 457 grams.

e)

Yes we can use the confidence interval to conduct the test in (c).

95% confidence interval for the true mean weight of all bags of coffee beans is (454.81, 459.19)

Confidence interval does not contain null value 454.

So we reject null hypothesis H0.

Conclusion: There is sufficient evidence to conclude that the true mean weight of all bags of coffee beans differs from 454 grams.