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Sagot :
Let's proceed with clarity and detail through each step of the hypothesis test.
### Step 1: Stating the Hypotheses
The sponsors want to test the hypothesis that children watch television at most [tex]\(20\)[/tex] hours per week.
- The null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu \leq 20\)[/tex]
- The alternative hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu > 20\)[/tex]
### Step 2: Calculating the Standard Error
The standard error (SE) is calculated using the formula:
[tex]\[ \text{SE} = \frac{\sigma}{\sqrt{n}} \][/tex]
Where:
- [tex]\(\sigma\)[/tex] is the population standard deviation ([tex]\(6\)[/tex] hours)
- [tex]\(n\)[/tex] is the sample size ([tex]\(30\)[/tex])
Plugging in the values:
[tex]\[ \text{SE} = \frac{6}{\sqrt{30}} \approx 1.0954 \][/tex]
### Step 3: Calculating the Z-Test Statistic
The z-test statistic is calculated using the formula:
[tex]\[ Z = \frac{\bar{x} - \mu_0}{\text{SE}} \][/tex]
Where:
- [tex]\(\bar{x}\)[/tex] is the sample mean ([tex]\(20.2667\)[/tex])
- [tex]\(\mu_0\)[/tex] is the hypothesized mean ([tex]\(20\)[/tex])
- [tex]\(\text{SE}\)[/tex] is the standard error ([tex]\(1.0954\)[/tex])
Plugging in the values:
[tex]\[ Z = \frac{20.2667 - 20}{1.0954} \approx 0.2435 \][/tex]
### Step 4: Calculating the P-value
Given the Z-test value, the next step is to find the corresponding p-value. In this case, because this is a one-tailed test where we're testing if the mean is greater than the hypothesized mean, we take:
[tex]\[ \text{P-value} = 1 - \text{CDF}(Z) \][/tex]
For [tex]\( Z = 0.2435 \)[/tex]:
[tex]\[ \text{P-value} \approx 0.4038 \][/tex]
### Step 5: Making the Decision
We compare the p-value with the significance level ([tex]\(\alpha = 0.1\)[/tex]):
- If [tex]\( \text{P-value} \leq \alpha \)[/tex], we reject [tex]\( H_0 \)[/tex].
- If [tex]\( \text{P-value} > \alpha \)[/tex], we fail to reject [tex]\( H_0 \)[/tex].
Here,
[tex]\[ \text{P-value} = 0.4038 \][/tex]
[tex]\[ \alpha = 0.1 \][/tex]
Since [tex]\(0.4038 > 0.1\)[/tex], we fail to reject the null hypothesis [tex]\( H_0 \)[/tex]. This means there is insufficient evidence to conclude that children watch more than [tex]\( 20 \)[/tex] hours of television per week.
### Summary of Results
1. Alternative Hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu > 20\)[/tex]
2. Standard Error: [tex]\( 1.0954 \)[/tex]
3. Test Statistic (Z-Test value): [tex]\( 0.2435 \)[/tex]
4. P-value: [tex]\( 0.4038 \)[/tex]
5. Conclusion: Fail to reject [tex]\( H_0 \)[/tex]. We accept the null hypothesis.
### Step 1: Stating the Hypotheses
The sponsors want to test the hypothesis that children watch television at most [tex]\(20\)[/tex] hours per week.
- The null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu \leq 20\)[/tex]
- The alternative hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu > 20\)[/tex]
### Step 2: Calculating the Standard Error
The standard error (SE) is calculated using the formula:
[tex]\[ \text{SE} = \frac{\sigma}{\sqrt{n}} \][/tex]
Where:
- [tex]\(\sigma\)[/tex] is the population standard deviation ([tex]\(6\)[/tex] hours)
- [tex]\(n\)[/tex] is the sample size ([tex]\(30\)[/tex])
Plugging in the values:
[tex]\[ \text{SE} = \frac{6}{\sqrt{30}} \approx 1.0954 \][/tex]
### Step 3: Calculating the Z-Test Statistic
The z-test statistic is calculated using the formula:
[tex]\[ Z = \frac{\bar{x} - \mu_0}{\text{SE}} \][/tex]
Where:
- [tex]\(\bar{x}\)[/tex] is the sample mean ([tex]\(20.2667\)[/tex])
- [tex]\(\mu_0\)[/tex] is the hypothesized mean ([tex]\(20\)[/tex])
- [tex]\(\text{SE}\)[/tex] is the standard error ([tex]\(1.0954\)[/tex])
Plugging in the values:
[tex]\[ Z = \frac{20.2667 - 20}{1.0954} \approx 0.2435 \][/tex]
### Step 4: Calculating the P-value
Given the Z-test value, the next step is to find the corresponding p-value. In this case, because this is a one-tailed test where we're testing if the mean is greater than the hypothesized mean, we take:
[tex]\[ \text{P-value} = 1 - \text{CDF}(Z) \][/tex]
For [tex]\( Z = 0.2435 \)[/tex]:
[tex]\[ \text{P-value} \approx 0.4038 \][/tex]
### Step 5: Making the Decision
We compare the p-value with the significance level ([tex]\(\alpha = 0.1\)[/tex]):
- If [tex]\( \text{P-value} \leq \alpha \)[/tex], we reject [tex]\( H_0 \)[/tex].
- If [tex]\( \text{P-value} > \alpha \)[/tex], we fail to reject [tex]\( H_0 \)[/tex].
Here,
[tex]\[ \text{P-value} = 0.4038 \][/tex]
[tex]\[ \alpha = 0.1 \][/tex]
Since [tex]\(0.4038 > 0.1\)[/tex], we fail to reject the null hypothesis [tex]\( H_0 \)[/tex]. This means there is insufficient evidence to conclude that children watch more than [tex]\( 20 \)[/tex] hours of television per week.
### Summary of Results
1. Alternative Hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu > 20\)[/tex]
2. Standard Error: [tex]\( 1.0954 \)[/tex]
3. Test Statistic (Z-Test value): [tex]\( 0.2435 \)[/tex]
4. P-value: [tex]\( 0.4038 \)[/tex]
5. Conclusion: Fail to reject [tex]\( H_0 \)[/tex]. We accept the null hypothesis.
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