The hypothesis test shows that we reject the null hypothesis and there is sufficient evidence to support the claim that the return rate is less than 20%
The claim that the return rate is less than 20% is p < 0.2. From the given information, we can compute our null hypothesis and alternative hypothesis as:
[tex]\mathbf{H_o :p =0.2}[/tex]
[tex]\mathbf{H_i:p < 0.2}[/tex]
Given that:
Sample size (n) = 6965
Sample proportion [tex]\mathbf{\hat p = \dfrac{x}{n} = \dfrac{1302}{6965} \sim0.1869}[/tex]
The test statistics for this data can be computed as:
[tex]\mathbf{z = \dfrac{\hat p - p}{\sqrt{\dfrac{p(1-p)}{n}}}}[/tex]
[tex]\mathbf{z = \dfrac{0.1869 -0.2}{\sqrt{\dfrac{0.2(1-0.2)}{6965}}}}[/tex]
[tex]\mathbf{z = \dfrac{-0.0131}{0.0047929}}[/tex]
z = -2.73
From the hypothesis testing, since the p < alternative hypothesis, then our test is a left-tailed test(one-tailed.
Hence, the p-value for the test statistics can be computed as:
P-value = P(Z ≤ z)
P-value = P(Z ≤ - 2.73)
By using the Excel function =NORMDIST (-2.73)
P-value = 0.00317
P-value ≅ 0.003
Therefore, we can conclude that since P-value is less than the significance level at ∝ = 0.01, we reject the null hypothesis and there is sufficient evidence to support the claim that the return rate is less than 20%
Learn more about hypothesis testing here:
https://brainly.com/question/15980493
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