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 p 1 be the proportion of customer complaints with the old dough and p 2 be the proportion of customer complaints with the new dough. State the competing hypotheses to determine if the proportion of customer complaints using the old dough is less than the proportion of customer complaints using the new dough g

Since the p value is a very low value and using any significance level 5% or 10% we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of customer complaints using the old dough is less than the proportion of customer complaints using the new dough.

Step-by-step explanation:

Information provided

[tex]X_{1}=6[/tex] represent the complaints with the old dough

[tex]X_{2}=16[/tex] represent the complaints with the new dough

[tex]n_{1}=385[/tex] sample 1 selected

[tex]n_{2}=340[/tex] sample 2 selected

[tex]p_{1}=\frac{6}{385}=0.0156[/tex] represent the proportion of complaints with the old dough

[tex]p_{2}=\frac{16}{340}=0.0471[/tex] represent the proportion of complaints with the new dough

[tex]\hat p[/tex] represent the pooled estimate of p

z would represent the statistic

[tex]p_v[/tex] represent the value

Hypothesis to test

We want to verify if the proportion of customer complaints using the old dough is less than the proportion of customer complaints using the new dough, the system of hypothesis would be:

Null hypothesis:[tex]p_{1} \geq p_{2}[/tex]

Alternative hypothesis:[tex]p_{1} < p_{2}[/tex]

The statitsic is given by:

[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex] (1)

Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{6+16}{385+340}=0.0303[/tex]

Replacing the info provided we got:

[tex]z=\frac{0.0156-0.0471}{\sqrt{0.0303(1-0.0303)(\frac{1}{385}+\frac{1}{340})}}=-2.469[/tex]

Now we can calculate the p value with this probability:

[tex]p_v =P(Z<-2.469)= 0.0068[/tex]

Since the p value is a very low value and using any significance level 5% or 10% we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of customer complaints using the old dough is less than the proportion of customer complaints using the new dough.