2. Herpetologists (snake specialists) found that a certain species of reticulated python have an average length of 20.5 feet with a standard deviation of 2.3 feet. The scientists collect a random sample of 30 adult pythons and measure their lengths. In their sample the mean length was 19.5 feet long. One of the herpetologists fears that pollution might be affecting the natural growth of the pythons. Do you think this sample result is unusually small?
Let's set up a null and alternative hypothesis, find a formula, then go from there.
Null hypothesis:
Ho: µ = 20.5 -->meaning: population mean is equal to 20.5
Ha: µ < 20.5 -->meaning: population mean is less than 20.5
Using the z-test formula to find the test statistic:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
z = (19.5 - 20.5)/(2.3/√30)
Calculate the z-test statistic and draw your conclusions. Remember if the null is not rejected, there is no difference. If the null is rejected, there is a difference.
I hope this will help.
To determine whether the sample result of 19.5 feet is unusually small, we can use hypothesis testing. In this case, we want to test whether the sample mean length of 19.5 feet is significantly different from the population mean length of 20.5 feet.
First, we need to define the null hypothesis (H0) and alternative hypothesis (Ha):
H0: The average length of the pythons in the population is 20.5 feet.
Ha: The average length of the pythons in the population is not 20.5 feet.
We can perform a one-sample t-test to evaluate the null hypothesis. The formula for the t-test statistic is:
t = (sample mean - population mean) / (standard deviation / sqrt(sample size))
Given that the sample mean is 19.5 feet, the population mean is 20.5 feet, the standard deviation is 2.3 feet, and the sample size is 30, we can calculate the t-value:
t = (19.5 - 20.5) / (2.3 / sqrt(30))
Simplifying further:
t = -1 / (2.3 / sqrt(30))
Using a calculator or software, we can determine the value of t. Let's assume the calculated t-value is -2.438.
Next, we need to determine the critical value based on the significance level (α) chosen for the test. For example, if we use a significance level of 0.05 (or 5%), we would use a two-tailed test.
Using either a t-distribution table or software, we can find the critical t-values for a two-tailed test with a significance level of 0.05. Let's assume the critical t-values are -2.048 and 2.048.
Finally, we compare the calculated t-value with the critical t-values to make a decision.
If the calculated t-value falls outside the range between the critical t-values, we reject the null hypothesis. This means the sample result is considered unusually small, indicating a significant difference between the sample mean length and the population mean length.
However, if the calculated t-value falls within the range between the critical t-values, we fail to reject the null hypothesis. This suggests that the sample result is not significantly different from the population mean length, and therefore, the sample result is not considered unusually small.
Note: The actual values of the t-value, critical t-values, and the decision may vary based on the calculations. This is just an example of how to approach the question using hypothesis testing.
To determine whether the sample result is unusually small, we need to compare it to the expected population mean length of 20.5 feet. We can use a statistical test called the z-test to assess the significance of the difference between the sample mean and the population mean.
The z-test allows us to calculate a z-score, which tells us how many standard deviations the sample mean is away from the population mean. The formula for calculating the z-score is:
z = (x - μ) / (σ / √n)
Where:
- x is the sample mean (19.5 feet in this case)
- μ is the population mean (20.5 feet)
- σ is the standard deviation (2.3 feet)
- n is the sample size (30)
We can calculate the z-score as follows:
z = (19.5 - 20.5) / (2.3 / √30)
z ≈ -1.62 / (2.3 / 5.48) ≈ -1.62 / 0.42 ≈ -3.86
To interpret the z-score, we can refer to a standard normal distribution table or use a calculator to determine the corresponding p-value. The p-value represents the probability of getting a sample mean as extreme as the one observed (or more extreme) if the population mean were actually 20.5 feet.
A small p-value (usually less than 0.05) indicates that the observed sample mean is significantly different from the population mean, suggesting that the sample result is unusually small. Conversely, a large p-value suggests that the observed difference could be due to random sampling variability alone, and it is not statistically significant.
Given that the z-score is approximately -3.86, it corresponds to an extremely small p-value, indicating that the observed sample mean of 19.5 feet is significantly different from the expected population mean of 20.5 feet. Therefore, based on this analysis, we can conclude that the sample result is unusually small, suggesting the possibility of pollution affecting the natural growth of the pythons.