Paul owns a mobile wood-fired pizza oven operation. A couple of his clients complained about his dough at a recent catering, so he changed his dough to a newer product. Using the old dough, there were 6 complaints out of 385 pizzas. With the new dough, there were 16 complaints out of 340 pizzas. Let p1 be the proportion of customer complaints with the old dough and p2 be the proportion of customer complains with the new dough. Based on a 95% confidence for the difference of the proportions, what can be concluded? Multiple ChoiceA. Reject H0, we can conclude the proportion of customer complaints is more for the old doughB. Do not reject H0, we cannot conclude the proportion of customer complaints is more for the old doughC. Do not reject H0, we can conclude the proportion of customer complaints is more for the old doughD. Reject H0, we cannot conclude the proportion of customer complaints is more for the old dough

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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