a perfume company claims that the best selling perfume contains at most 25% alcohol. twenty bottles were selected and found to have mean of 29.7% and standard deviation of 4.8%. test the claim of the perfume company at the 0.0& level pf significance. to test the null.hypothesis, the best selling perfume contains at most 25% alcohol.
It would help if you proofread your questions before you posted them.
Level of significance?
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to t he Z score.
25
Well, it seems like the perfume company claimed that their best selling perfume contains at most 25% alcohol. Let's put their claim to the test!
To do this, we will use a hypothesis test with the following hypotheses:
- Null hypothesis (H0): The best selling perfume contains at most 25% alcohol.
- Alternative hypothesis (H1): The best selling perfume contains more than 25% alcohol.
Next, we need to calculate the test statistic. In this case, we will be using a one-sample t-test since we have a sample mean and standard deviation.
The formula for the t-test statistic is:
t = (sample mean - claimed mean) / (sample standard deviation / sqrt(sample size))
Substituting the values we have:
t = (29.7 - 25) / (4.8 / sqrt(20))
Calculating this, we find t to be approximately 4.06.
Now, we need to compare this t-value with the critical t-value at a 0.05 level of significance. For a one-tailed test with 19 degrees of freedom, the critical t-value is approximately 1.73.
Since our t-value (4.06) is greater than the critical t-value (1.73), we have strong evidence to reject the null hypothesis. Therefore, we can conclude that the best selling perfume contains more than 25% alcohol, contrary to the company's claim.
I hope my statistical humor helped lighten the mood while discussing hypothesis testing!
To test the claim of the perfume company that the best selling perfume contains at most 25% alcohol, we will conduct a hypothesis test at a significance level of 0.05.
Here are the steps to perform the hypothesis test:
Step 1: State the null hypothesis (H0) and the alternative hypothesis (Ha):
- Null hypothesis (H0): The best selling perfume contains at most 25% alcohol.
- Alternative hypothesis (Ha): The best selling perfume contains more than 25% alcohol.
Step 2: Select the appropriate test statistic:
Since we are testing the mean of a sample, we will use the t-test with a one-sample design.
Step 3: Set up the decision rule:
We will reject the null hypothesis if the calculated test statistic is larger than the critical value from the t-distribution with (n-1) degrees of freedom and a significance level of 0.05.
Step 4: Calculate the test statistic and p-value:
The formula for the t-test statistic for a one-sample design is:
t = (x̄ - μ) / (s / √n)
Where:
x̄ = sample mean
μ = hypothesized population mean
s = sample standard deviation
n = sample size
In this case, x̄ = 29.7%, μ = 25%, s = 4.8%, and n = 20.
t = (29.7 - 25) / (4.8 / √20)
t = 4.7 / (4.8 / √20)
t = 6.65
Using statistical software or a t-table, we find that the critical value for a one-tailed test with 19 degrees of freedom and a significance level of 0.05 is approximately 1.729.
Step 5: Make a decision:
Since the calculated t-value (6.65) is greater than the critical value (1.729), we reject the null hypothesis.
Step 6: Interpret the result:
The data provides sufficient evidence to conclude that the best selling perfume contains more than 25% alcohol.
To test the claim of the perfume company, we can use a hypothesis test. Let's define our null hypothesis (H0) and alternative hypothesis (Ha) as follows:
H0: The best selling perfume contains at most 25% alcohol.
Ha: The best selling perfume contains more than 25% alcohol. (This is the opposite of H0.)
Next, we need to determine the appropriate test statistic to use. Since we have a sample mean and standard deviation, and we want to test the mean against a claimed value, we can use a one-sample t-test.
To perform the test, we'll follow these steps:
Step 1: Set up the null and alternative hypotheses.
- H0: μ ≤ 25% (The true mean is less than or equal to 25%.)
- Ha: μ > 25% (The true mean is greater than 25%.)
Step 2: Determine the significance level.
- The problem states a significance level of 0.05 (or 5%).
Step 3: Compute the test statistic.
- The test statistic for a one-sample t-test is given by: t = (x̄ - μ₀) / (s / √n)
- x̄ : the sample mean (29.7%)
- μ₀ : the claimed value (25%)
- s : the sample standard deviation (4.8%)
- n : the sample size (20)
Calculating the test statistic:
t = (29.7 - 25) / (4.8 / √20)
t ≈ 2.622
Step 4: Determine the critical value(s).
- Since we're testing the claim that the true mean is greater than 25%, this is a right-tailed test.
- With a significance level of 0.05, the critical value for a one-tailed test with 19 degrees of freedom is approximately 1.729 (from t-table or calculator).
Step 5: Make a decision.
- If the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Since 2.622 > 1.729, we reject the null hypothesis.
Step 6: Interpret the decision.
- Based on the sample data, there is sufficient evidence to conclude that the true mean alcohol content of the best selling perfume is greater than 25% at the 0.05 level of significance.
Therefore, the claim made by the perfume company is not supported by the sample data.