To solve this hypothesis testing problem, let's go through the steps:

Step 1: Define the Hypotheses

Based on the problem, the engineer wants to test if employees at start-up companies work more hours on average than the national average of 47 hours. This indicates a one-tailed test.

1. Null Hypothesis (H₀): The mean work week at start-ups is equal to or less than 47 hours.
   • $H_0: \mu = 47$
2. Alternative Hypothesis (H₁): The mean work week at start-ups is more than 47 hours.
   • $H_1: \mu > 47$

Step 2: Calculate the Sample Mean and Standard Deviation

The data provided:

• Hours: 49, 48, 60, 50, 49, 68, 57, 60, 49, 46, 55, 50

Using this dataTo solve this hypothesis testing problem, let's go through the steps:

Step 1: Define the Hypotheses

Based on the problem, the engineer wants to test if employees at start-up companies work more hours on average than the national average of 47 hours. This indicates a one-tailed test.

1. Null Hypothesis (H₀): The mean work week at start-ups is equal to or less than 47 hours.
   • $H_0: \mu = 47$
2. Alternative Hypothesis (H₁): The mean work week at start-ups is more than 47 hours.
   • $H_1: \mu > 47$

Step 2: Calculate the Sample Mean and Standard Deviation

The data provided:

• Hours: 49, 48, 60, 50, 49, 68, 57, 60, 49, 46, 55, 50

Using this

1. Calculate the sample mean ( $\bar{x}$ ).
2. Calculate the sample standard deviation ( $s$ ).

Step 3: Compute the Test Statistic

Since the population standard deviation is not given, we'll use the t-test statistic for a one-sample test: $t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$ where:

• $\bar{x}$ = sample mean,
• $\mu = 47$ (hypothesized population mean),
• $s$ = sample standard deviation,
• $n = 12$ (sample size).

Step 4: Find the p-value and Make a Decision

With a 5% significance level ( $\alpha = 0.05$ ), we’ll compare the computed t-statistic against the critical value from the t-distribution with $n - 1 = 11$ degrees of freedom.

If $p$-value $< 0.05$, we reject the null hypothesis.

I'll now perform these calculations.Here are the calculated values for the hypothesis test:

1. Sample Mean ( $\bar{x}$ ): 53.4167 hours
2. Sample Standard Deviation ( $s$ ): 6.6121 hours
3. Test Statistic ( $t$ ): 3.3617
4. p-value: 0.0032

Decision

Since the p-value (0.0032) is less than the significance level of 0.05, we reject the null hypothesis.

Conclusion

The test supports the claim that the mean number of hours worked by employees at start-ups is greater than the US mean of 47 hours.

Would you like more details on any part of the process?

Related Questions:

1. What is the difference between a one-tailed and two-tailed test?
2. Why do we use the t-test instead of the z-test in this scenario?
3. How does the p-value help in decision-making?
4. What would be the implications of a Type I error in this context?
5. How would results change if we used a 1% significance level instead?

Tip:

When performing hypothesis testing, always check if the test is one-tailed or two-tailed, as it affects the interpretation of the p-value.