Answer: B. Do not reject H0, we cannot conclude the proportion of customer complaints is more for the old dough
Step-by-step explanation:
This is a test of 2 population proportions. Let 1 and 2 be the subscript for the old and the new dough players. The population proportions of customer complaints with the old and new dough would be p1 and p2
P1 - P2 = difference in the proportion of customer complaints with the old and new dough.
The null hypothesis is
H0 : p1 ≥ p2
pm - pw ≥ 0
The alternative hypothesis is
Ha : p1 < p2
p1 - p2 < 0
it is a left tailed test
Sample proportion = x/n
Where
x represents number of success(number of complaints)
n represents number of samples
For old dough
x1 = 6
n1 = 385
P1 = 6/385 = 0.016
For new dough,
x2 = 16
n2 = 340
P2 = 16/340 = 0.047
The pooled proportion, pc is
pc = (x1 + x2)/(n1 + n2)
pc = (6 + 16)/(385 + 340) = 0.03
1 - pc = 1 - 0.03 = 0.97
z = (Pm - Pw)/√pc(1 - pc)(1/nm + 1/nw)
z = (0.016 - 0.047)/√(0.03)(0.97)(1/385 + 1/340) = - 0.031/√0.00553857907
z = - 0.42
Since it is a left tailed test, we would find the p value for the area to the left of the z score. From the normal distribution table,
p value = 0.337
For a 95% confidence level, the significant level, alpha is
1 - 0.95 = 0.05
Since 0.05 < 0.337, we would accept the null hypothesis
Therefore, Do not reject H0, we cannot conclude the proportion of customer complaints is more for the old dough
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