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The z-test is a statistical test used when the standard deviation of the population is known, and the sample size is sufficiently large. In this scenario, we have a detergent whose viscosity we want to test against a known average. Let's simplify this.

To decide whether the average viscosity of the detergent is 800 centistokes, we set up a statistical test. We perform the z-test using the formula:

• \[ z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \]

Here:

• \( \bar{x} \) is the sample mean (812 in this case)
• \( \mu_0 \) is the hypothesized population mean (800)
• \( \sigma \) is the population standard deviation (25)
• \( n \) is the sample size (16)

The calculated z-value is 1.92, which indicates how far the sample mean is from the population mean in terms of standard error units.

We compare the calculated z-value with the critical values (for a two-tailed test with \( \alpha = 0.05 \)), which are \( \pm 1.96 \). The calculated z-value (1.92) does not fall outside this range, leading us to conclude that there is not enough evidence to reject the null hypothesis.

Confidence intervals provide a range of values that likely contain the true population mean. They give insights into the precision of our mean estimate from the sample data.

To calculate a 95% confidence interval for the population mean, we use the formula:

• \[ \bar{x} \pm z_{critical} \times \frac{\sigma}{\sqrt{n}} \]

Let's break it down:

• \( \bar{x} \) is the sample mean (812)
• \( z_{critical} \) is the critical value for 95% confidence, typically 1.96 for a normal distribution
• \( \sigma \) is the population standard deviation (25)
• \( n \) is the sample size (16)

Plugging these values into the formula gives us a range from 799.75 to 824.25. This means we are 95% confident that the true mean viscosity of the detergent lies within this interval.

This interval supports the hypothesis test's results because the hypothesized mean of 800 centistokes is within this confidence range, suggesting it is plausible.

The P-value represents the probability of obtaining test results at least as extreme as the observed results, under the null hypothesis.

In the context of the hypothesis test, the P-value helps us decide whether to reject the null hypothesis. It is found by calculating how likely the observed z-value (1.92) would be, assuming the null hypothesis is true.

From the standard normal distribution table, a z-value of 1.92 has a corresponding P-value of about 0.054. Because this P-value is greater than our chosen significance level \( \alpha = 0.05 \), it suggests our sample data is not sufficiently extreme to reject the null hypothesis.

Therefore, with a P-value of 0.054, there is a 5.4% chance of observing such an extreme value if the null hypothesis were indeed true. This means the result doesn't provide strong enough evidence to discard the null. It highlights the role of P-values in making informed decisions about statistical hypotheses.