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%
<h3>What is the claim that the return rate is less than 20% by using a statistical hypothesis method?</h3>
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:


Given that:
Sample size (n) = 6965
Sample proportion 
The test statistics for this data can be computed as:



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:
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Answer:
First, second, third, and fourth are functions
Step-by-step explanation:
Each input in the first, second, third, and fourth only have one output. In the fifth one 3.3 and 6.6 are seen twice as the x-value, so this is incorrect.
(answered this on your other question, but I will put it here too!)
Cada entrada en la primera, segunda, tercera y cuarta tiene solo una salida. En el quinto, 3.3 y 6.6 se ven dos veces como el valor de x, por lo que esto es incorrecto.
(respondí esto en tu otra pregunta, ¡pero también lo pondré aquí!)
Answer: 72 HOPE THIS HELPS!
33 = 88x
x = 33/88
x = .375 = 37.5 percent