One study claimed that 86 % of college students identify themselves as procrastinators. A professor believes that the claim regarding college students is too high. The professor conducts a simple random sample of 139 college students and finds that 110 of them identify themselves as procrastinators. Does this evidence support the professor's claim that fewer than 86% of college students are procrastinators? Use a 0.05 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision.
One study claimed that 86 % of college students identify themselves as procrastinators. A professor believes that the claim regarding college students is too high. The professor conducts a simple random sample of 139 college students and finds that 110 of them identify themselves as procrastinators. Does this evidence support the professor's claim that fewer than 86% of college students are procrastinators? Use a 0.05 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision.
MATLAB: An Introduction with Applications
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![### Statistical Analysis of Procrastination Among College Students
#### Claim and Hypothesis
One study reported that **86%** of college students identify themselves as procrastinators. A professor challenges this claim, suspecting that the percentage might be lower. To test this hypothesis, the professor conducts a simple random sample consisting of **139 college students**, finding that **110** of them identify as procrastinators.
#### Objective
Determine whether the professor's sample supports the hypothesis that fewer than **86%** of college students are procrastinators, utilizing a **0.05 level of significance**.
#### Steps to Conduct the Hypothesis Test:
**Step 3 of 3: Draw a conclusion and interpret the decision.**
This step involves concluding if the sample evidence provides sufficient basis to reject the initial claim.
#### Conclusion Drawing and Interpretation:
1. **Estimate Population Proportion**: Calculate the sample proportion (p̂) of procrastinators:
\[
\hat{p} = \frac{110}{139} \approx 0.791
\]
2. **Formulate Hypotheses**:
- Null Hypothesis (H₀): \( p \geq 0.86 \)
- Alternative Hypothesis (H₁): \( p < 0.86 \)
3. **Calculate Test Statistic**: Apply the z-test for proportion:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} = \frac{0.791 - 0.86}{\sqrt{\frac{0.86 \cdot (1 - 0.86)}{139}}}
\]
4. **Determine Critical Value and p-value**:
- Critical value for a one-tailed test at 0.05 significance: \( z_{0.05} \approx -1.645 \)
- Compare calculated z-value with this critical value or calculate the p-value for z.
5. **Make a Decision**:
- If z calculated > critical z, fail to reject H₀.
- If z calculated < critical z, reject H₀.
6. **Interpret the Decision**:
- Based on the above comparison, derive whether the evidence strongly counters or supports the professor's claim. A decision to reject H₀ would indicate evidence supporting that fewer](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d050742-a776-4688-997d-9b380febe959%2F8311dbf7-6e5e-4de1-9c96-c1990f1cf41c%2Fod12xy_processed.png&w=3840&q=75)
Transcribed Image Text:### Statistical Analysis of Procrastination Among College Students
#### Claim and Hypothesis
One study reported that **86%** of college students identify themselves as procrastinators. A professor challenges this claim, suspecting that the percentage might be lower. To test this hypothesis, the professor conducts a simple random sample consisting of **139 college students**, finding that **110** of them identify as procrastinators.
#### Objective
Determine whether the professor's sample supports the hypothesis that fewer than **86%** of college students are procrastinators, utilizing a **0.05 level of significance**.
#### Steps to Conduct the Hypothesis Test:
**Step 3 of 3: Draw a conclusion and interpret the decision.**
This step involves concluding if the sample evidence provides sufficient basis to reject the initial claim.
#### Conclusion Drawing and Interpretation:
1. **Estimate Population Proportion**: Calculate the sample proportion (p̂) of procrastinators:
\[
\hat{p} = \frac{110}{139} \approx 0.791
\]
2. **Formulate Hypotheses**:
- Null Hypothesis (H₀): \( p \geq 0.86 \)
- Alternative Hypothesis (H₁): \( p < 0.86 \)
3. **Calculate Test Statistic**: Apply the z-test for proportion:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} = \frac{0.791 - 0.86}{\sqrt{\frac{0.86 \cdot (1 - 0.86)}{139}}}
\]
4. **Determine Critical Value and p-value**:
- Critical value for a one-tailed test at 0.05 significance: \( z_{0.05} \approx -1.645 \)
- Compare calculated z-value with this critical value or calculate the p-value for z.
5. **Make a Decision**:
- If z calculated > critical z, fail to reject H₀.
- If z calculated < critical z, reject H₀.
6. **Interpret the Decision**:
- Based on the above comparison, derive whether the evidence strongly counters or supports the professor's claim. A decision to reject H₀ would indicate evidence supporting that fewer
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