Making sure that the scales used by businesses in the United States are accurate is the responsibility of the National Institute for Standards and Technology (NIST) in Washington, D.C. Suppose that NIST technicians are testing a scale by using a weight known to weigh exactly grams. The standard deviation for scale reading is known to be o=2. They weigh this weight on the scale 47 times and read the result each time. The 47 scale readings have a sample mean of x=999.1 grams. The calibration point is set too low if the mean scale reading is less than 1000 grams. The technicians want to perform a hypothesis test to determine whether the calibration point is set too low. Use the a=0.01 level of significance and the P-value method with the TI-84 Plus calculator.

Question

(b) Compute the value of the test statistic. Round the answer to two decimal places.

Answer 1

To compute the value of the test statistic for the given hypothesis test, we will use the formula for the z-test for a single mean. The formula is:

\[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]

Where:

• \(\bar{x}\) is the sample mean (999.1 grams),
• \(\mu\) is the population mean under the null hypothesis (1000 grams),
• \(\sigma\) is the known population standard deviation (2 grams),
• \(n\) is the sample size (47).

Now we can plug in the values:

\[ z = \frac{999.1 - 1000}{2 / \sqrt{47}} \]

First, we calculate the denominator:

\[ 2 / \sqrt{47} \approx 2 / 6.855654 (approximately \sqrt{47} \approx 6.855654) \] \[ 2 / \sqrt{47} \approx 0.2919 \]

Next, we compute the numerator:

\[ 999.1 - 1000 = -0.9 \]

Now we can calculate \(z\):

\[ z = \frac{-0.9}{0.2919} \approx -3.08 \]

So, rounding to two decimal places, the value of the test statistic is:

\[ \boxed{-3.08} \]

Answer 2

(c) Find the p-value. Round the answer to four decimal places.

Answer 3

To find the p-value for the z-test statistic calculated previously, we use the standard normal distribution. Since this is a left-tailed test (the null hypothesis states that the mean is 1000 grams, and we are testing whether it is less than that), we need to find the area to the left of the calculated z-value.

The calculated z-value is approximately \(-3.08\). We can look up this z-value in the standard normal distribution table or use a calculator.

On the TI-84 Plus calculator, you can find the p-value using the following steps:

1. Press the 2nd button.
2. Press VARS to access the distribution menu.
3. Select normalcdf(.
4. Input the values for the left tail up to your z-value:

\[ \text{normalcdf}(-\infty, -3.08) \]

This can be approximated as:

\[ \text{normalcdf}(-1E99, -3.08) \]

5. Input should look like this: normalcdf(-1E99, -3.08)

When you execute this, the calculator will provide the p-value.

Using the Z-table, or through a standard calculator, we find:

The p-value for \(z = -3.08\) is approximately \(0.0010\).

Thus, the p-value rounded to four decimal places is:

\[ \boxed{0.0010} \]

Answer 4

We know we reject Hv0, is there enough evidence to conclude anything

Answer 5

Since you have decided to reject the null hypothesis \(H_0: \mu = 1000\) grams at the \(\alpha = 0.01\) significance level, and given that you have calculated a p-value of 0.0010, which is considerably less than the significance level of 0.01, it provides strong evidence against the null hypothesis.

Here’s what this means:

1. Rejection of \(H_0\): Rejecting the null hypothesis means that you have found sufficient evidence to conclude that the mean scale reading is less than 1000 grams.

2. Conclusion: Based on the test results, you can conclude that the calibration point for the scale is set too low, indicating that the scale needs to be adjusted. This conclusion is statistically significant given the p-value and the selected significance level.

Overall, you have enough evidence to state that there is strong support for the claim that the calibration of the scale is inaccurate and that the mean weight measured by the scale is indeed less than the known weight of 1000 grams.