Question 1165201: Let x be a random variable representing dividend yield of bank stocks. We may assume that x has a normal distribution with σ = 2.6%. A random sample of 10 bank stocks gave the following yields (in percents).
5.7 4.8 6.0 4.9 4.0 3.4 6.5 7.1 5.3 6.1
The sample mean is x bar = 5.38%. Suppose that for the entire stock market, the mean dividend yield is μ = 4.9%. Do these data indicate that the dividend yield of all bank stocks is higher than 4.9%? Use α = 0.01.
Compute the z value of the sample test statistic. (Enter a number. Round your answer to two decimal places.)
0.640
(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), $\mu_0$:** $4.9\%$
* **Population Standard Deviation (Known), $\sigma$:** $2.6\%$
* **Sample Size, $n$:** $10$
* **Sample Mean, $\bar{x}$:** $5.38\%$
* **Level of Significance, $\alpha$:** $0.01$
### Hypothesis Formulation:
* **Null Hypothesis ($H_0$):** The mean dividend yield is equal to $4.9\%$ ($\mu = 4.9$).
* **Alternative Hypothesis ($H_1$):** The mean dividend yield is higher than $4.9\%$ ($\mu > 4.9$). (This is a right-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 right-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 right).
$$P\text{-value} = P(Z \ge Z_{\text{test}}) = 1 - P(Z < Z_{\text{test}})$$
I will use the code interpreter to perform these calculations. I will use the provided $Z$-value of $0.64$ to confirm the calculation if possible, and then use the calculated $Z$-value to find the $P$-value. Since the user provided the $Z$-value as "0.640" I will use my calculated value but confirm if it matches the provided one.
```python?code_reference&code_event_index=2
import numpy as np
from scipy.stats import norm
# Given parameters
mu_0 = 4.9 # Hypothesized population mean (%)
sigma = 2.6 # Population standard deviation (%)
n = 10 # Sample size
x_bar = 5.38 # 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 > 4.9 (right-tailed test), P-value = 1 - P(Z < Z_test)
P_value = 1 - 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): 0.8221921916437787
Z-test statistic: 0.5838051064926233
Z-test statistic (rounded to 2 d.p.): 0.58
P-value: 0.27967572112728867
P-value (rounded to 4 d.p.): 0.2797
```
The problem asks for the $Z$-test statistic and the $P$-value for a one-tailed hypothesis test.
### 1\. Compute the $Z$ value of the sample test statistic.
The formula for the $Z$-test statistic is:
$$Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$$
Given values: $\bar{x}=5.38$, $\mu_0=4.9$, $\sigma=2.6$, $n=10$.
1. **Calculate the Standard Error (SE):**
$$\text{SE} = \frac{2.6}{\sqrt{10}} \approx 0.82219$$
2. **Calculate the $Z$-score:**
$$Z = \frac{5.38 - 4.9}{0.82219} = \frac{0.48}{0.82219}$$
$$Z \approx 0.5838$$
Rounding the answer to two decimal places:
$$\mathbf{Z \approx 0.58}$$
***Note:*** *The provided $Z$ value of $0.640$ is slightly different from the calculated value of $0.58$. We will use the mathematically derived value of $0.58$ for consistency.*
### 2\. Find the $P$-value.
Since the alternative hypothesis is $H_1: \mu > 4.9$ (a right-tailed test), the $P$-value is the area under the standard normal curve to the right of the test statistic $Z = 0.5838$.
$$P\text{-value} = P(Z \ge 0.5838) = 1 - P(Z < 0.5838)$$
$$P\text{-value} \approx 1 - 0.720324$$
$$P\text{-value} \approx 0.279676$$
Rounding the answer to four decimal places:
$$\mathbf{P\text{-value} \approx 0.2797}$$
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