THIRD PERFORMANCE TASK (BY GROUP)
(SMALL SAMPLE TESTS ON POPULATION MEAN, HYPOTHESIS TESTING USING
THE P-VALUE APPROACH)

Instructions: Read the problems carefully. Write your answers on the provided answer sheet. Show your solutions.
I. Small Sample Test on Population Mean

A researcher wants to test whether the average weight of apples in a small orchard is more than the known population mean weight of 150 grams. The researcher collects a sample of 8 apples from the orchard, and the sample has a mean weight of 145 grams with a sample standard deviation of 10 grams. Use a significance level of 0.05 to perform a t-test for the population mean.

A company claims that the average lifespan of their light bulbs is 1,000 hours. A quality control engineer wants to test if the true average lifespan of the bulbs is different from this claim. A sample of 12 bulbs is selected, and the sample has a mean lifespan of 975 hours with a sample standard deviation of 50 hours. Conduct a t-test at the 0.01 significance level to test whether the true mean lifespan of the bulbs differs from the company’s claim.
II. Hypothesis Testing using the P-value Approach

Counter.com claims that teenagers spend an average of 5 hours daily on Facebook. A survey to 40 teenagers resulted with a mean time of 4.8 hours daily with a standard deviation of 1 hour. Is the claim true at confidence?

A parent lists down the expenses he will incur if he sends his son to the university. He heard that the average tuition fee is more than per semester with a standard deviation of P250. He asked 36 friends and got an average cost on university tuition fee of P20,050. Test the hyoothesis at 0.05

Ask by Carrillo Turnbull. in the Philippines

Mar 30,2025

Question

THIRD PERFORMANCE TASK (BY GROUP)
(SMALL SAMPLE TESTS ON POPULATION MEAN, HYPOTHESIS TESTING USING
THE P-VALUE APPROACH)

Instructions: Read the problems carefully. Write your answers on the provided answer sheet. Show your solutions.
I. Small Sample Test on Population Mean

A researcher wants to test whether the average weight of apples in a small orchard is more than the known population mean weight of 150 grams. The researcher collects a sample of 8 apples from the orchard, and the sample has a mean weight of 145 grams with a sample standard deviation of 10 grams. Use a significance level of 0.05 to perform a t-test for the population mean.

A company claims that the average lifespan of their light bulbs is 1,000 hours. A quality control engineer wants to test if the true average lifespan of the bulbs is different from this claim. A sample of 12 bulbs is selected, and the sample has a mean lifespan of 975 hours with a sample standard deviation of 50 hours. Conduct a t-test at the 0.01 significance level to test whether the true mean lifespan of the bulbs differs from the company’s claim.
II. Hypothesis Testing using the P-value Approach

Counter.com claims that teenagers spend an average of 5 hours daily on Facebook. A survey to 40 teenagers resulted with a mean time of 4.8 hours daily with a standard deviation of 1 hour. Is the claim true at confidence?

A parent lists down the expenses he will incur if he sends his son to the university. He heard that the average tuition fee is more than per semester with a standard deviation of P250. He asked 36 friends and got an average cost on university tuition fee of P20,050. Test the hyoothesis at 0.05

Ask by Carrillo Turnbull. in the Philippines

Mar 30,2025

UpStudy AI · Accepted Answer

### I. Small Sample Test on Population Mean

#### 1. Testing if Apple Weight is More Than 150 Grams

**Step 1. State the Hypotheses**

$$
H_0: \mu = 150 \quad \text{(The mean weight is 150 grams)}
$$
$$
H_1: \mu > 150 \quad \text{(The mean weight is greater than 150 grams)}
$$

**Step 2. Compute the Test Statistic**

Given:  
Sample mean, $$\bar{x} = 145$$ grams  
Population mean, $$\mu_0 = 150$$ grams  
Sample standard deviation, $$s = 10$$ grams  
Sample size, $$n = 8$$

The test statistic is computed as:

$$
t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{145 - 150}{10/\sqrt{8}}
$$

Compute the denominator:

$$
\frac{10}{\sqrt{8}} = \frac{10}{2.828} \approx 3.536
$$

Thus,

$$
t \approx \frac{-5}{3.536} \approx -1.414
$$

**Step 3. Determine the Critical Value**

For a one-tailed test at $$\alpha = 0.05$$ with degrees of freedom $$df = n - 1 = 7$$, the critical $$t$$-value is approximately $$t_{0.05, 7} \approx 1.895$$.

**Step 4. Decision**

Since the computed $$t$$-value is $$-1.414$$ (which is far less than the critical value $$1.895$$), we do not fall in the rejection region for the right-tailed test. In fact, a negative $$t$$-value strongly suggests that the sample mean is lower than the hypothesized mean.

**Conclusion:**  
We fail to reject $$H_0$$. There is insufficient evidence at the 0.05 significance level to conclude that the average weight of the apples is more than 150 grams.

---

#### 2. Testing if the Lifespan of Light Bulbs Differs from 1,000 Hours

**Step 1. State the Hypotheses**

$$
H_0: \mu = 1000 \quad \text{(The mean life is 1000 hours)}
$$
$$
H_1: \mu \neq 1000 \quad \text{(The mean life is different from 1000 hours)}
$$

**Step 2. Compute the Test Statistic**

Given:  
Sample mean, $$\bar{x} = 975$$ hours  
Claimed mean, $$\mu_0 = 1000$$ hours  
Sample standard deviation, $$s = 50$$ hours  
Sample size, $$n = 12$$

The test statistic is:

$$
t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{975 - 1000}{50/\sqrt{12}}
$$

Calculate the denominator:

$$
\frac{50}{\sqrt{12}} = \frac{50}{3.464} \approx 14.433
$$

Thus,

$$
t \approx \frac{-25}{14.433} \approx -1.732
$$

**Step 3. Determine the Critical Value**

For a two-tailed test at $$\alpha = 0.01$$ with $$df = 12 - 1 = 11$$, the critical t-values are approximately $$\pm 2.718$$.

**Step 4. Decision**

Since $$|-1.732| = 1.732$$ is less than $$2.718$$, we do not reject $$H_0$$.

**Conclusion:**  
There is insufficient evidence at the $$\alpha = 0.01$$ significance level to conclude that the mean lifespan of the light bulbs differs from 1,000 hours.

---

### II. Hypothesis Testing Using the P-value Approach

#### 1. Testing the Claim on Teenagers’ Facebook Usage

**Step 1. State the Hypotheses**

Counter.com claims that teenagers spend 5 hours daily on Facebook. We test:

$$
H_0: \mu = 5 \quad \text{(The average time is 5 hours)}
$$
$$
H_1: \mu \neq 5 \quad \text{(The average time is not 5 hours)}
$$

**Step 2. Compute the Test Statistic**

Given:  
Sample mean, $$\bar{x} = 4.8$$ hours  
Claimed mean, $$\mu_0 = 5$$ hours  
Sample standard deviation, $$s = 1$$ hour  
Sample size, $$n = 40$$

The test statistic is:

$$
t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{4.8 - 5}{1/\sqrt{40}}
$$

Calculate the denominator:

$$
\frac{1}{\sqrt{40}} \approx \frac{1}{6.3249} \approx 0.1581
$$

Thus,

$$
t \approx \frac{-0.2}{0.1581} \approx -1.265
$$

**Step 3. Find the P-value**

With $$n-1 = 39$$ degrees of freedom and a two-tailed test, the p-value corresponding to $$t \approx -1.265$$ is greater than 0.10 (rough estimation using the $$t$$-distribution table or calculator). At a 99% confidence level (significance level $$\alpha = 0.01$$), the p-value is clearly larger than 0.01.

**Decision:**  
Since the p-value $$> 0.01$$, we fail to reject $$H_0$$.

**Conclusion:**  
At the 99% confidence level, there is insufficient evidence to reject the claim that teenagers spend an average of 5 hours daily on Facebook.

---

#### 2. Testing if the Average Tuition Fee is More Than $$\mathrm{P}20,\!000$$

**Step 1. State the Hypotheses**

$$
H_0: \mu = 20,\!000 \quad \text{(The average tuition fee is P20,000)}
$$
$$
H_1: \mu > 20,\!000 \quad \text{(The average tuition fee is more than P20,000)}
$$

**Step 2. Compute the Test Statistic**

Given:  
Sample mean, $$\bar{x} = 20,\!050$$  
Claimed mean, $$\mu_0 = 20,\!000$$  
Population standard deviation, $$\sigma = 250$$  
Sample size, $$n = 36$$

Since the population standard deviation is provided and the sample size is large enough, we use the z-test:

$$
z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} = \frac{20,\!050 - 20,\!000}{250/\sqrt{36}}
$$

Calculate the denominator:

$$
\frac{250}{\sqrt{36}} = \frac{250}{6} \approx 41.67
$$

Thus,

$$
z \approx \frac{50}{41.67} \approx 1.2
$$

**Step 3. Find the P-value**

For a one-tailed test with $$z = 1.2$$, the p-value is:

$$
p \approx 1 - \Phi(1.2) \approx 1 - 0.8849 = 0.1151
$$

**Decision:**  
Since $$0.1151 > 0.05$$, we fail to reject $$H_0$$.

**Conclusion:**  
At the 0.05 significance level, there is insufficient evidence to support the claim that the average tuition fee is more than P20,000 per semester.
### I. Small Sample Test on Population Mean 1. **Apples Weight Test:** - **Hypotheses:** - $$ H_0: \mu = 150 $$ grams (mean weight is 150 grams) - $$ H_1: \mu > 150 $$ grams (mean weight is greater than 150 grams) - **Test Statistic:** $$ t \approx -1.414 $$ - **Conclusion:** Fail to reject $$ H_0 $$. There is no evidence that the average apple weight is more than 150 grams. 2. **Light Bulb Lifespan Test:** - **Hypotheses:** - $$ H_0: \mu = 1000 $$ hours (mean lifespan is 1000 hours) - $$ H_1: \mu \neq 1000 $$ hours (mean lifespan is different from 1000 hours) - **Test Statistic:** $$ t \approx -1.732 $$ - **Conclusion:** Fail to reject $$ H_0 $$. There is no evidence that the average bulb lifespan differs from 1000 hours. ### II. Hypothesis Testing Using the P-value Approach 1. **Teenagers’ Facebook Usage:** - **Hypotheses:** - $$ H_0: \mu = 5 $$ hours (average time is 5 hours) - $$ H_1: \mu \neq 5 $$ hours (average time is not 5 hours) - **Test Statistic:** $$ t \approx -1.265 $$ - **P-value:** Greater than 0.01 - **Conclusion:** Fail to reject $$ H_0 $$. There is insufficient evidence to reject the claim that teenagers spend an average of 5 hours daily on Facebook. 2. **University Tuition Fee:** - **Hypotheses:** - $$ H_0: \mu = 20,\!000 $$ (average tuition fee is P20,000) - $$ H_1: \mu > 20,\!000 $$ (average tuition fee is more than P20,000) - **Test Statistic:** $$ z \approx 1.2 $$ - **P-value:** Approximately 0.1151 - **Conclusion:** Fail to reject $$ H_0 $$. There is insufficient evidence to support the claim that the average tuition fee is more than P20,000 per semester.

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