Period =2r /b
assume frequency factor meaning angular frequency "b"
then
2r= 2r/b => b=1
A. sine function whose period 2pi radians
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:
brainly.com/question/15980493
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Answer:
34347
Step-by-step explanation:
Use the formula s=p(1+r/k)^kt
p=30000
r=7% (interest)
k=1 (because of per annum)
t= 2 (years)
Hence: 30000(1+0.07/1)^1(2) = 34347
Answer:
f = 5/6 = 0.833 or f = 0
Step-by-step explanation: