To solve this hypothesis testing problem, let’s go through each part step-by-step.

1. Setting up the Hypotheses

We have the following information:

• The population mean ($\mu$) of work hours for the U.S. is 47 hours.
• The sample data consists of hours worked by employees at start-up companies.

We want to test if employees at start-ups work more hours than the U.S. average, so our hypotheses are:

• Null Hypothesis (H₀): $\mu = 47$
• Alternative Hypothesis (H₁): $\mu > 47$

This will be a one-tailed test since we are only interested in whether the mean is greater than 47 hours.

2. Calculating the Sample Mean and Standard Deviation

Let’s calculate the mean and standard deviation of the sample. The data provided is:

$45, 48, 56, 51, 45, 54, 55, 58, 49, 49, 49, 54$

Sample Mean (x̄) and Sample Standard Deviation (s) can be calculated using these values.

• Mean (x̄): $x̄ = \frac{\sum X}{n} = \frac{45 + 48 + 56 + 51 + 45 + 54 + 55 + 58 + 49 + 49 + 49 + 54}{12}$

• Sample Standard Deviation (s): $s = \sqrt{\frac{\sum (X - x̄)^2}{n - 1}}$

I will calculate these values next.

3. Test Statistic

We use the t-test because the population standard deviation is unknown and the sample size is relatively small ($n = 12$).

The t-statistic formula is: $t = \frac{x̄ - \mu}{\frac{s}{\sqrt{n}}}$ where:

• $x̄$ = sample mean
• $\mu$ = population mean (47 hours)
• $s$ = sample standard deviation
• $n$ = sample size (12)

4. Finding the p-value

With a 1% significance level ($\alpha = 0.01$), we will compare the calculated t-value to the critical value from the t-distribution table with $n - 1 = 11$ degrees of freedom.

If the p-value is less than 0.01, we reject the null hypothesis. Otherwise, we fail to reject it.

Let’s calculate each of these values.