Question 1165134: 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.)
(c) Find (or estimate) the P-value.
Answer by CPhill(2138)   (Show Source):
You can put this solution on YOUR website! ```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"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
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
```
This is a one-sample $Z$-test for the mean.
### 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:
$$\mathbf{Z \approx 0.58}$$
-----
### 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 $Z = 0.5838$.
$$P\text{-value} = P(Z \ge 0.5838)$$
Using the standard normal distribution calculation:
$$P\text{-value} \approx 0.279676$$
Rounding the answer to four decimal places:
$$\mathbf{P\text{-value} \approx 0.2797}$$
-----
### Conclusion (Optional)
Since the $P$-value $(0.2797)$ is much greater than the significance level $\alpha$ $(0.01)$, we fail to reject the null hypothesis. The data do not provide sufficient evidence to conclude that the dividend yield of all bank stocks is higher than $4.9\%$.
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