(b)	A professor claims that graduate students spend more time on research compared to undergraduate students. The mean number of hours spent on research per week for undergraduate students is 20 hours with a standard deviation of 3 hours per week. A sample of 50 graduate students was selected at random, and the mean number of hours spent on research per week was found to be 22 hours. Perform the required hypothesis test at 5% significance level to check whether the professor’s claim can be accepted or not.
		(8 marks)


To test the professor's claim that graduate students spend more time on research compared to undergraduate students, we will conduct a hypothesis test for the difference in means. 

### Step-by-step Solution:

#### 1. **Formulate the Hypotheses:**

The null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)) are:

- \(H_0\): The mean time spent on research by graduate students is equal to or less than that of undergraduate students.
  
  \[
  H_0: \mu_{\text{graduate}} \leq \mu_{\text{undergraduate}} = 20
  \]

- \(H_1\): The mean time spent on research by graduate students is greater than that of undergraduate students.

  \[
  H_1: \mu_{\text{graduate}} > 20
  \]

This is a **one-tailed test**.

#### 2. **Given Data:**

- Mean research hours for undergraduate students: \(\mu_{\text{undergraduate}} = 20\)
- Standard deviation of research hours for undergraduate students: \(\sigma_{\text{undergraduate}} = 3\)
- Sample size of graduate students: \(n = 50\)
- Sample mean of research hours for graduate students: \(\bar{x} = 22\)
- Significance level: \(\alpha = 0.05\)

#### 3. **Determine the Test Statistic:**

We use a **one-sample z-test** since the standard deviation is known, and the sample size is large (\(n > 30\)).

The test statistic is calculated as:

\[
z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
\]

Substituting the values:

\[
z = \frac{22 - 20}{\frac{3}{\sqrt{50}}}
\]

\[
z = \frac{2}{\frac{3}{7.071}}
\]

\[
z = \frac{2}{0.4243}
\]

\[
z \approx 4.71
\]

#### 4. **Find the Critical Value and Make a Decision:**

For a one-tailed test at the 5% significance level (\(\alpha = 0.05\)), the critical value of \(z\) is approximately 1.645.

- If \(z > 1.645\), we reject the null hypothesis \(H_0\).
- If \(z \leq 1.645\), we fail to reject the null hypothesis \(H_0\).

Since our calculated \(z \approx 4.71\) is much greater than 1.645, we **reject the null hypothesis**.

#### 5. **Conclusion:**

At the 5% significance level, we have sufficient evidence to reject the null hypothesis. This means we **accept the alternative hypothesis** that the mean time spent on research by graduate students is significantly greater than that of undergraduate students. Therefore, the professor's claim is supported by the data.

### Would you like more details or have any questions?

Here are some related questions that might interest you:
1. How would the hypothesis test change if the sample size was smaller (e.g., less than 30)?
2. What is the meaning of a p-value in the context of hypothesis testing?
3. How do you perform a hypothesis test when the population standard deviation is unknown?
4. What are Type I and Type II errors in hypothesis testing, and what are their implications in this context?
5. How can the power of a hypothesis test be increased?

**Tip:** When conducting hypothesis tests, always check the assumptions of normality and the sample size requirements to ensure the test's validity.

Explore the hypothesis test comparing research hours of graduate and undergraduate students. Includes step-by-step solution using the z-test formula with specific data points: mean, standard deviation, sample size, and significance level.

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