In an effort to make children’s toys safer and more tamperproof, toy packaging has become cumbersome for parents to remove in many cases. Accordingly, the director of marketing at Toys4Tots, a large toy manufacturer, wants to evaluate the effectiveness of a new packaging design that engineers claim will reduce customer complaints by more than 10 percentage points. Customer satisfaction surveys were sent to 220 parents who registered toys packaged under the old design and 220 parents who registered toys packaged under the new design. Of these, 86 parents expressed dissatisfaction with packaging of the old design, and 41 parents expressed dissatisfaction with packaging of the new design.
Let p1 represent the population proportion of parents that expressed dissatisfaction with the packaging of the old design and p2 represent the population proportion of parents that expressed dissatisfaction with the packaging of the new design.
A. Specify the null and alternative hypotheses to test for whether the proportion of customer complaints has decreased by more than 5% under the new packaging design.
1. H0 : p1 - p2 < 0.10; HA : p1 - p2 > 0.10
2. H0 : p1 - p2 = 0.10; HA : p1 - p2 is not equal to 0.10
3. H0 : p1 - p2 > 0.10; HA : p1 - p2 < 0.10
B. What is the value of the test statistic and the associated p-value?
C. At the 1% significance level, do the results support the engineers' claim?
D. At the 10% significance level, do the results support the engineers' claim?

Answer:

Explained below.

Step-by-step explanation:

(a)

In this case we need to determine if the proportion of customer complaints has decreased by more than 10% under the new packaging design.

The hypothesis can be defined as follows:  

H₀: The proportion of customer complaints has not decreased by more than 10% under the new packaging design, i.e. p₁ - p₂ ≤ 0.10.  

Hₐ: The proportion of customer complaints has decreased by more than 10% under the new packaging design, i.e. p₁ - p₂ > 0.10.  

(b)

The information provided is:

n₁ = 220

n₂ = 220

X₁ = 86

X₂ = 41

Compute the sample proportions and total proportions as follows:

[tex]\hat p_{1}=\frac{X_{1}}{n_{1}}=\frac{86}{220}=0.3909\\\\\hat p_{2}=\frac{X_{2}}{n_{2}}=\frac{41}{220}=0.1864\\\\\hat P=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{86+41}{220+220}=0.2886[/tex]

Compute the test statistic value as follows:

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

   [tex]=\frac{0.3909-0.1864}{\sqrt{0.2886(1-0.2886)\times[\frac{1}{220}+\frac{1}{220}]}}\\\\=\frac{0.2045}{\sqrt{0.0018665}}\\\\=4.733524\\\\\approx 4.73[/tex]

The test statistic value is 4.73.

Compute the p-value as follows:

[tex]p-value=P(Z>4.73)[/tex]

               [tex]=1-P(Z<4.73)\\=1-(\approx 1)\\=0[/tex]

The p-value of the test is 0.

(c)

The decision rule is:

The null hypothesis will be rejected if the p-value of the test is less than the significance level.

p-value = 0 < α = 0.01.

The null hypothesis will be rejected at 1%significance level.

Thus, there is enough evidence at 1% significance level to support the engineers' claim.

(d)

The decision rule is:

The null hypothesis will be rejected if the p-value of the test is less than the significance level.

p-value = 0 < α = 0.10.

The null hypothesis will be rejected at 10%significance level.

Thus, there is enough evidence at 10% significance level to support the engineers' claim.

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