Question 1165202: Bill Alther is a zoologist who studies Anna's hummingbird (Calypte anna). (Reference: Hummingbirds, K. Long, W. Alther.) Suppose that in a remote part of the Grand Canyon, a random sample of six of these birds was caught, weighed, and released. The weights (in grams) were as follows.
3.7 2.9 3.8 4.2 4.8 3.1
The sample mean is x bar = 3.75 grams. Let x be a random variable representing weights of hummingbirds in this part of the Grand Canyon. We assume that x has a normal distribution and σ = 0.64 gram. Suppose it is known that for the population of all Anna's hummingbirds, the mean weight is μ = 4.30 grams. Do the data indicate that the mean weight of these birds in this part of the Grand Canyon is less than 4.30 grams? Use α = 0.10
Compute the z value of the sample test statistic. (Enter a number. Round your answer to two decimal places.)
(c) Find (or estimate) the P-value. (Enter a number. Round your answer to four decimal places.)
Answer by CPhill(2138)   (Show Source):
You can put this solution on YOUR website! The user is asking for the calculation of the $Z$-test statistic and the corresponding $P$-value for a one-sample hypothesis test concerning a population mean ($\mu$), where the population standard deviation ($\sigma$) is known.
### Given Information:
* **Population Mean (Hypothesized), $\mu_0$:** $4.30$ grams
* **Population Standard Deviation (Known), $\sigma$:** $0.64$ grams
* **Sample Size, $n$:** $6$
* **Sample Mean, $\bar{x}$:** $3.75$ grams
* **Level of Significance, $\alpha$:** $0.10$
### Hypothesis Formulation:
* **Null Hypothesis ($H_0$):** The mean weight is equal to $4.30$ grams ($\mu = 4.30$).
* **Alternative Hypothesis ($H_1$):** The mean weight is less than $4.30$ grams ($\mu < 4.30$). (This is a left-tailed test.)
### (a) Compute the $Z$ value of the sample test statistic.
The formula for the $Z$-test statistic for a sample mean when $\sigma$ is known is:
$$Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$$
### (c) Find the $P$-value.
Since this is a left-tailed test, the $P$-value is the probability of observing a $Z$-score as extreme as the calculated test statistic or more extreme (i.e., further to the left).
$$P\text{-value} = P(Z \le Z_{\text{test}})$$
I will use the code interpreter to perform these calculations.
```python?code_reference&code_event_index=2
import numpy as np
from scipy.stats import norm
# Given parameters
mu_0 = 4.30 # Hypothesized population mean
sigma = 0.64 # Population standard deviation
n = 6 # Sample size
x_bar = 3.75 # Sample mean
# --- Part (a): Compute the Z value of the sample test statistic ---
# 1. Calculate the standard error of the mean
SE = sigma / np.sqrt(n)
# 2. Calculate the Z-score
Z_test = (x_bar - mu_0) / SE
# --- Part (c): Find the P-value ---
# Since H1 is mu < 4.30 (left-tailed test), P-value = P(Z <= Z_test)
P_value = norm.cdf(Z_test)
# Output results rounded as requested
Z_rounded = round(Z_test, 2)
P_rounded = round(P_value, 4)
print(f"Standard Error (SE): {SE}")
print(f"Z-test statistic: {Z_test}")
print(f"Z-test statistic (rounded to 2 d.p.): {Z_rounded}")
print(f"P-value: {P_value}")
print(f"P-value (rounded to 4 d.p.): {P_rounded}")
```
```text?code_stdout&code_event_index=2
Standard Error (SE): 0.26127890589687236
Z-test statistic: -2.105030247704293
Z-test statistic (rounded to 2 d.p.): -2.11
P-value: 0.017644337582746063
P-value (rounded to 4 d.p.): 0.0176
```
Here are the required calculations for the hypothesis test:
### (a) Compute the $Z$ value of the sample test statistic.
The $Z$-test statistic is calculated as:
$$Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$$
Using the given values ($\bar{x}=3.75$, $\mu_0=4.30$, $\sigma=0.64$, $n=6$):
1. **Calculate the Standard Error (SE):**
$$\text{SE} = \frac{0.64}{\sqrt{6}} \approx 0.26128$$
2. **Calculate the $Z$-score:**
$$Z = \frac{3.75 - 4.30}{0.26128} = \frac{-0.55}{0.26128}$$
$$Z \approx -2.1050$$
Rounding the answer to two decimal places:
$$\mathbf{Z \approx -2.11}$$
### (c) Find the $P$-value.
Since the alternative hypothesis is $H_1: \mu < 4.30$ (a left-tailed test), the $P$-value is the area under the standard normal curve to the left of the test statistic $Z = -2.1050$.
$$P\text{-value} = P(Z \le -2.1050)$$
$$P\text{-value} \approx 0.017644$$
Rounding the answer to four decimal places:
$$\mathbf{P\text{-value} \approx 0.0176}$$
-----
*(For context: Since the $P$-value $(0.0176)$ is less than the significance level $\alpha$ $(0.10)$, we would reject the null hypothesis and conclude that the data indicates the mean weight of these birds in this part of the Grand Canyon is indeed less than $4.30$ grams.)*
|
|
|
