a manufacturer of detergent claims that the contents of the boxes in the market weigh on average at least 16 oz.. the distribution of weights is known to be normal with a standard deviation of 0.4 oz. to test the claim of the manufacturer a random sample of 16 boxes were weighed and it was found that the sample mean weight is 15.84 oz. test at levels of significance of 5%, 10% and the null hypothesis that the population mean weight is at least 16 oz.

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 proportions (.05 and .10) for one-tailed test related to the Z score.

To test the claim of the manufacturer regarding the average weight of the detergent boxes, you can perform a hypothesis test.

The null hypothesis (H0) states that the population mean weight is at least 16 oz. The alternative hypothesis (H1) would be that the population mean weight is less than 16 oz.

Now, let's calculate the test statistics and p-values for the given sample:

Step 1: Set up the hypotheses:
H0: μ ≥ 16 (null hypothesis)
H1: μ < 16 (alternative hypothesis)

Step 2: Determine the significance level:
The significance levels given are 5% and 10%, so we will conduct the test at these two levels.

Step 3: Calculate the test statistic:
The formula for the test statistic is:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the population mean, s is the standard deviation, and n is the sample size.

For this problem, x̄ = 15.84 oz, μ = 16 oz, s = 0.4 oz, and n = 16.

t = (15.84 - 16) / (0.4 / √16)
= (-0.16) / (0.4 / 4)
= -0.16 / 0.1
= -1.6

Step 4: Determine the critical value or p-value:
To find the critical value or p-value, we need to use the t-distribution table or calculator.

At a significance level of 5% (α = 0.05), the critical t-value for a one-tailed test with 16 degrees of freedom is approximately -1.746.

At a significance level of 10% (α = 0.10), the critical t-value for a one-tailed test with 16 degrees of freedom is approximately -1.340.

Step 5: Make a decision:
- If the test statistic is less than the critical value, reject the null hypothesis.
- If the p-value is less than the significance level, reject the null hypothesis.

Now, let's compare the test statistic with the critical value and p-value for both significance levels:

For α = 0.05:
Since the test statistic (-1.6) is greater than the critical value (-1.746), we fail to reject the null hypothesis.

For α = 0.10:
Since the test statistic (-1.6) is greater than the critical value (-1.340), we fail to reject the null hypothesis.

Thus, at both the 5% and 10% levels of significance, we fail to reject the null hypothesis. This means that there is not enough evidence to suggest that the average weight of the detergent boxes is less than 16 oz.

To test the claim of the manufacturer, we will conduct a hypothesis test using the given sample data.

The null hypothesis (H0) is that the population mean weight is at least 16 oz.
The alternative hypothesis (Ha) is that the population mean weight is less than 16 oz.

We'll use a t-test since the population standard deviation is unknown.

The steps to perform the hypothesis test are as follows:

Step 1: State the hypotheses
- Null hypothesis (H0): μ >= 16 (population mean weight is at least 16 oz.)
- Alternative hypothesis (Ha): μ < 16 (population mean weight is less than 16 oz.)

Step 2: Set the significance level
- Significance level (α) is given as 5% and 10%.

Step 3: Compute the test statistic
- Compute the test statistic t using the formula:
t = (sample mean - hypothesized mean) / (sample standard deviation / √n)
- Where the sample mean is 15.84 oz, hypothesized mean is 16 oz, sample standard deviation is 0.4 oz, and n (sample size) is 16.

Step 4: Determine the critical value(s)
- The critical value(s) depends on the significance level and the degrees of freedom.
- Since the sample size is 16, the degrees of freedom (df) = n - 1 = 16 - 1 = 15.
- Look up the critical value(s) for the desired significance level and degrees of freedom in the t-distribution table.

Step 5: Make a decision
- If the test statistic t is less than the critical value(s), reject the null hypothesis.
- If the test statistic t is greater than or equal to the critical value(s), fail to reject the null hypothesis.

Step 6: Draw a conclusion
- Based on the decision made in step 5, we conclude whether there is enough evidence to support the manufacturer's claim or not.

Now, let's calculate the test statistic t and determine the critical value(s) for the given significance levels of 5% and 10%:

Step 1: State the hypotheses:
- Null hypothesis (H0): μ >= 16
- Alternative hypothesis (Ha): μ < 16

Step 2: Set the significance level:
- Significance level (α) = 5%, 10%

Step 3: Compute the test statistic:
- Sample mean (x̄) = 15.84 oz
- Hypothesized mean (μ0) = 16 oz
- Sample standard deviation (s) = 0.4 oz
- Sample size (n) = 16

t = (15.84 - 16) / (0.4 / √16)

Step 4: Determine the critical value(s):
- Degrees of freedom (df) = n - 1 = 16 - 1 = 15
- t-critical for α = 0.05, df = 15: -1.761
- t-critical for α = 0.10, df = 15: -1.341

Step 5: Make a decision:
- Compare the calculated t-value with the critical value(s) from step 4.

--- For α = 0.05 ---
- Calculated t-value = t = (15.84 - 16) / (0.4 / √16)
= -1.999
- Since -1.999 < -1.761, we reject the null hypothesis.

--- For α = 0.10 ---
- Calculated t-value = t = (15.84 - 16) / (0.4 / √16)
= -1.999
- Since -1.999 < -1.341, we reject the null hypothesis.

Step 6: Draw a conclusion:
- Based on the results, there is enough evidence to conclude that the population mean weight is less than 16 oz at both the 5% and 10% significance levels. Thus, we reject the manufacturer's claim.