To test 

Upper H 0 : sigma equals 70H0: σ=70 

versus 

Upper H 1 : sigma less than 70H1: σ<70​, 

a random sample of size 

n equals 22n=22 

is obtained from a population that is known to be normally distributed.

​(a) If the sample standard deviation is determined to be 

s equals 61.3s=61.3​, 

compute the test statistic.

​(b) If the researcher decides to test this hypothesis at the 

alpha equals 0.05α=0.05 

level of​ significance, use  technology to determine the​ P-value.

​(c) Will the researcher reject the null​ hypothesis?

Question content area bottom

Part 1

​(a) The test statistic is 

chi Subscript 0 Superscript 2χ20equals=16.1016.10.

​(Round to two decimal places as​ needed.)

Part 2

​(b) The​ P-value is 

enter your response here.

​(Round to three decimal places as​ needed.)

To solve this hypothesis testing problem, we'll address each part systematically.

---

### **(a) Compute the test statistic**

The formula for the test statistic in a chi-square test for variance is:

\[
\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}
\]

where:  
- \( n = 22 \) (sample size)  
- \( s = 61.3 \) (sample standard deviation)  
- \( \sigma_0 = 70 \) (hypothesized population standard deviation)  

Substitute values into the formula:

1. Compute \( s^2 \) (sample variance):  
   \[
   s^2 = 61.3^2 = 3757.69
   \]

2. Compute the test statistic:  
   \[
   \chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{(22-1)(3757.69)}{70^2}
   \]
   \[
   \chi^2 = \frac{21 \times 3757.69}{4900} = \frac{78811.49}{4900} \approx 16.10
   \]

So, the test statistic is:
\[
\chi^2 \approx 16.10
\]

---

### **(b) Determine the P-value**

For a one-tailed chi-square test, the P-value is calculated from the chi-square distribution with \( df = n - 1 \). Here:
- \( df = 22 - 1 = 21 \)
- Test statistic: \( \chi^2 = 16.10 \)

The P-value corresponds to the area in the left tail of the chi-square distribution (since \( H_1 \) is \( \sigma < 70 \)). Using technology or a chi-square table:

\[
P(\chi^2 < 16.10, df = 21)
\]

Using statistical software or a calculator, the P-value is approximately:

\[
P \approx 0.372
\]

---

### **(c) Decision at \( \alpha = 0.05 \)**

- The null hypothesis \( H_0: \sigma = 70 \) is rejected if the P-value \( < \alpha \).  
- Here, \( P = 0.372 \) is much larger than \( \alpha = 0.05 \).

Thus, **the researcher does not reject the null hypothesis**.

---

### Final Summary:
(a) Test statistic: \( \chi^2 = 16.10 \)  
(b) P-value: \( P \approx 0.372 \)  
(c) Decision: Do not reject \( H_0 \).  

---

Would you like more details or have additional questions?  

### **Expansion Questions**:
1. How is the chi-square test used in other variance testing scenarios?
2. How does the chi-square distribution shape change with degrees of freedom?
3. What are the assumptions underlying a chi-square test?
4. How would the conclusion change if \( \alpha = 0.10 \)?
5. How can technology (e.g., software like Excel or Python) help in hypothesis testing?

**Tip**: Always verify that your sample is from a normally distributed population before using chi-square tests for variance.

To test H₀: σ = 70 versus H₁: σ < 70, a random sample of size n = 22 is obtained from a population that is known to be normally distributed.

(a) If the sample standard deviation is determined to be s = 61.3, compute the test statistic.

(b) If the researcher decides to test this hypothesis at the α = 0.05 level of significance, use technology to determine the P-value.

(c) Will the researcher reject the null hypothesis?

Learn how to perform a chi-square test for variance with this detailed example. Compute the test statistic, find the P-value, and make a decision on rejecting the null hypothesis for a sample size of n = 22, sample standard deviation s = 61.3, and significance level α = 0.05.

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