91Ó°ÊÓ

Gentle Ben is a Morgan horse at a Colorado dude ranch. Over the past 8 weeks, a veterinarian took the following glucose readings from this horse (in \(\mathrm{mg} / 100 \mathrm{ml}\) ). \(\begin{array}{llllllll}93 & 88 & 82 & 105 & 99 & 110 & 84 & 89\end{array}\) The sample mean is \(\bar{x}=93.8\). Let \(x\) be a random variable representing glucose readings taken from Gentle Ben. We may assume that \(x\) has a normal distribution, and we know from past experience that \(\sigma=12.5\). The mean glucose level for horses should be \(\mu=85 \mathrm{mg} / 100 \mathrm{ml}\) (Reference: Merck Veterinary Mamul). Do these data indicate that Gentle Ben has an overall average glucose level higher than 85 ? Use \(\alpha=0.05\).

Step by step solution

01

State the Hypotheses

To determine if Gentle Ben has an average glucose level different from 85 mg/100 ml, we set up the hypotheses. The null hypothesis (0) states that the mean glucose level  = 85 mg/100 ml. The alternative hypothesis (1) posits that the mean glucose level  > 85 mg/100 ml.

02

Determine the Test Statistic

Use the formula for the test statistic: \( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \). Plug in the values: \( \bar{x} = 93.8, \mu = 85, \sigma = 12.5, n = 8 \). Therefore, \( z = \frac{93.8 - 85}{\frac{12.5}{\sqrt{8}}} \).

03

Calculate the Test Statistic

Calculate the standard error: \( \frac{12.5}{\sqrt{8}} \approx 4.42 \). Then calculate the z-score: \( z = \frac{93.8 - 85}{4.42} \approx 2.00 \).

04

Determine the P-Value

Using a standard normal distribution table, find the p-value for \( z = 2.00 \). The p-value corresponding to a z-score of 2.00 is about 0.0228.

05

Make the Decision

Compare the p-value to the significance level \( \alpha = 0.05 \). Since 0.0228 < 0.05, we reject the null hypothesis.

06

Draw a Conclusion

Since the null hypothesis is rejected, we conclude that the data indicate that Gentle Ben's average glucose level is significantly higher than 85 mg/100 ml.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis

In hypothesis testing, the null hypothesis is like the default assumption that there is no effect or no difference. It's what we try to gather evidence against when conducting a test.
For this exercise, the null hypothesis ( H_0 ) states that Gentle Ben's average glucose level is 85 mg/100 ml. This implies there's nothing unusual about his glucose levels compared to the standard for horses.

• The null hypothesis is generally the statement you want to test.
• If your data shows sufficient evidence, the null hypothesis can be rejected.
• It's important not just to state it but understand what it represents - no change or impact.

This hypothesis provides a benchmark to measure if Gentle Ben's glucose levels naturally vary around 85 mg/100 ml or significantly differ.

Alternative Hypothesis

The alternative hypothesis is what you aim to support through your data. It's the opposite stance to the null hypothesis. Here, the alternative hypothesis ( H_1 ) suggests that Gentle Ben's mean glucose level is greater than 85 mg/100 ml.
When you opt for an alternative hypothesis, you're looking for evidence of a change or a new effect. This is particularly significant in situations where you suspect a difference based on prior knowledge or theoretical expectations.

• In this exercise, as Ben's apparent higher glucose levels suggest, we consider that his levels exceed the norm.
• This hypothesis represents what you believe to be true, requiring evidence to support these beliefs fully.
• Rejecting the null hypothesis indirectly supports the alternative hypothesis.

Z-Score

The Z-Score is a crucial part of hypothesis testing. It measures the number of standard deviations a data point is from the mean of a set of data. In this case, we use the test statistic formula:\( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \).
This tells us how far Ben's average glucose reading (\bar{x} = 93.8) is from the expected mean glucose level (\mu = 85), given the standard deviation (\sigma = 12.5) and the number of samples (n = 8).

• The calculated Z-score of approximately 2.00 indicates that Ben's glucose mean is 2 standard deviations above the expected mean.
• The higher the absolute value of the Z-score, the further away the sample mean is from the population mean, suggesting a more significant finding.
• This Z-score allows comparison across different datasets, thanks to its standardization.

P-Value

The P-Value helps determine the significance of the results. It's the probability that the observed data would occur under the null hypothesis assumption. In our example, the p-value for the Z-score of 2.00 is approximately 0.0228.
This small value suggests that the probability of observing such an extreme or more extreme value if the null hypothesis were true is very low.

• P-values help decide whether to reject the null hypothesis.
• A low p-value (< α = 0.05) indicates strong evidence against the null hypothesis, so we reject it.
• In our exercise, since 0.0228 < 0.05, the results are considered statistically significant.

This p-value reflects the evidence that suggests Ben's glucose levels are higher than the norm.

Standard Error

The standard error (SE) quantifies the amount of variability in the sample mean. It gives an idea of how much the sample mean (\bar{x}) would fluctuate around the population mean (\mu). Calculated by dividing the standard deviation by the square root of the sample size:\( \frac{12.5}{\sqrt{8}} \approx 4.42 \).

• A smaller SE indicates that the sample mean is a more accurate estimate of the population mean.
• This particular SE value tells us about the sampling distribution width and Ben’s sample averages.
• It plays a crucial role in computing the Z-score, thereby impacting the hypothesis test.

In hypothesis testing, understanding the SE helps comprehend the potential error and reliability of your sample estimates.