Question

4. (from Q31 P. 594) Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5. a. (1 mark) Calculate the sample mean, ¯x and its standard error s/√ n. b. Create a 95% confidence interval for the mean weight of such bags. c. State H0 and Ha and calculate a p-value if we want to show that the true mean weight of all such bags is actually greater than 28.3 grams. d. State your conclusion, referring to both the confidence interval and the p-value

Answer 1

a)

sample mean, xbar = 28.9833
sample standard deviation, s = 0.3601

std.dev = s/sqrt(n)
= 0.3601/sqrt(6)
= 0.1470

b)

sample size, n = 6
degrees of freedom, df = n - 1 = 5

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.57

ME = tc * s/sqrt(n)
ME = 2.57 * 0.3601/sqrt(6)
ME = 0.378

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (28.9833 - 2.57 * 0.3601/sqrt(6) , 28.9833 + 2.57 * 0.3601/sqrt(6))
CI = (28.61 , 29.36)

b)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 28.3
Alternative Hypothesis, Ha: μ > 28.3

Rejection Region
This is right tailed test, for α = 0.05 and df = 5
Critical value of t is 2.015.
Hence reject H0 if t > 2.015

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (28.9833 - 28.3)/(0.3601/sqrt(6))
t = 4.648

P-value Approach
P-value = 0.0028
As P-value < 0.05, reject the null hypothesis.

as the confidenc einterval does noyt contain 28.3 so, reject H0