Question

1. A candy manufacturer selects mints at random from the production line and weighs them. For one week, the day shift weighed n₁ = 194 mints, and the night shift weighed n₂ = 162 mints. The numbers of these mints that weighed at most 21 grams was Y₁ = 28 for the day shift and Y₂ = 11 for the night shift. Let ?₁ and ?₂ denote the proportions of all mints that weigh at most 21 grams for the day and night shifts, respectively.
   1. Give point estimates of ?₁, ?₂, and ?₁ −?₂.
   2. Do that data present significant evidence to show that the nights shift gives fewer mints that weigh at most 21 grams than the day shift? Answer this question by constructing a 90% confidence interval for ?₁ −?₂·
   3. Answer the same question in (b) by performing a relevant test. Give the pvalue of the test. What does this p-value tell you?

Answer 1

a)

p1cap = X1/N1 = 28/194 = 0.1443
p1cap = X2/N2 = 11/162 = 0.0679

point estimate = 0.1443 - 0.0679 = 0.0764

b)

Here, , n1 = 194 , n2 = 162
p1cap = 0.1443 , p2cap = 0.0679

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.1443 * (1-0.1443)/194 + 0.0679*(1-0.0679)/162)
SE = 0.032

For 0.9 CI, z-value = 1.6449
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.1443 - 0.0679 - 1.6449*0.032, 0.1443 - 0.0679 + 1.6449*0.032)
CI = (0.0238 , 0.129)

yes, because confidence interval does not contain 0

c)

pcap = (X1 + X2)/(N1 + N2) = (28+11)/(194+162) = 0.1096

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 > p2

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.1443-0.0679)/sqrt(0.1096*(1-0.1096)*(1/194 + 1/162))
z = 2.3

P-value Approach
P-value = 0.0107
As P-value < 0.1, reject the null hypothesis.