58% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the
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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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