If it is claimed that the contents of soaps boxes sold weigh on average at least 20 ounces.
The distribution of weight is known to be normal;
We assume \sigma = 0.5σ=0.5 is known
n=30
sample mean weight = 18.5 ounces.
Test at the 10% significance level (\alpha = .10)(α=.10) the null hypothesis that the population mean weight is at least 20 ounces.
90% = mean ± 1.645 SEm
SEm = SD/√n
Insert the values and calculate.
Forget the above. I goofed. Try this.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
If it
is claimed that the contents
of soaps
boxes sold
weigh on average at least 20 ounces.
The distribution
of weight is known to be normal;
We
assume \sigma = 0.5σ=0.5 is known
n=30
sample
mean
weight =
18.5 ounces.
Test
at
the 10% significance level (\alpha = .10)(α=.10) the null
hypothesis that the population mean weight is at least 20
ounces.
To test the null hypothesis that the population mean weight is at least 20 ounces, we can use a one-sample z-test.
Here's how you can calculate the test statistic and make the hypothesis test:
1. Formulate the null and alternative hypotheses:
- Null hypothesis (H0): The population mean weight is 20 ounces or more.
- Alternative hypothesis (Ha): The population mean weight is less than 20 ounces.
2. Calculate the test statistic (z-score) using the sample mean, population standard deviation, and sample size:
- The formula for the z-score is: z = (x̄ - μ) / (σ / √n)
- Given: x̄ = 18.5 ounces, σ = 0.5 ounces, n = 30
- Compute: z = (18.5 - 20) / (0.5 / √30)
3. Determine the critical value for the given significance level (α = 0.10):
- Since the alternative hypothesis is one-tailed (less than), we need to find the critical z-value for a 10% left-tailed test. This corresponds to a cumulative probability of 0.10.
- Using a standard normal distribution table or a calculator, find the z-value that corresponds to a cumulative probability of 0.10.
4. Compare the test statistic with the critical value:
- If the test statistic is less than the critical value, reject the null hypothesis; otherwise, fail to reject the null hypothesis.
5. Interpret the results:
- If you reject the null hypothesis, it means there is evidence to suggest that the population mean weight is less than 20 ounces.
- If you fail to reject the null hypothesis, it means there is not enough evidence to conclude that the population mean weight is less than 20 ounces.
Note: Since this is a one-sample z-test and the population standard deviation (σ) is known, we can directly use the z-test. If the population standard deviation is unknown, we would use the t-test instead.
Please let me know if you would like the actual calculations for steps 2 and 3, or if you have any further questions.