Question

what are the correct hypotheses?
ho: select an answer ? hours
h1: select an answer ? hours
based on the hypotheses, find the following:
test statistic= (round to 2 decimal places)
p - value= (round to 4 decimal places)
the correct decision is to select an answer
the correct summary would be: select an answer that the mean numb hours of all employees at start - up companies work more than the us mean of 47 hours.
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Explanation:

To solve this hypothesis testing problem (likely about the mean number of work hours), we follow these steps:

Step 1: Define the Hypotheses

The problem compares the mean work hours at start - up companies to the US mean of 47 hours.

• Null Hypothesis (\(H_0\)): This is the statement of no difference or the status quo. So, \(H_0: \mu = 47\) (the mean work hours at start - up companies is equal to 47 hours).
• Alternative Hypothesis (\(H_1\)): From the context “work more than the US mean”, we have a right - tailed test. So, \(H_1: \mu>47\) (the mean work hours at start - up companies is greater than 47 hours).

Step 2: Test Statistic (Assuming a z - test or t - test, let's assume we have sample data. For example, if we have a sample mean \(\bar{x}\), sample standard deviation \(s\), sample size \(n\))

The formula for the z - test statistic (when population standard deviation \(\sigma\) is known) is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). For a t - test (when \(\sigma\) is unknown), the formula is \(t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\).

Suppose we have a sample: Let's say \(n = 54\) (from the top - left box), sample mean \(\bar{x}\) (let's assume we calculated it as, for example, 50), population mean \(\mu = 47\), and sample standard deviation \(s = 5\) (these are assumed values for illustration, as the problem's full data is not shown).

For a t - test:
\(t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}=\frac{50 - 47}{\frac{5}{\sqrt{54}}}=\frac{3}{\frac{5}{7.3485}}\approx\frac{3}{0.6804}\approx4.41\) (rounded to 2 decimal places).

Step 3: p - value

For a right - tailed t - test with \(df=n - 1=54 - 1 = 53\) and test statistic \(t = 4.41\), we can use a t - distribution table or a calculator. Using a calculator (like the t - distribution function in statistical software), the p - value is very small, say approximately \(0.0000\) (rounded to 4 decimal places).

Step 4: Decision Rule

We compare the p - value to a significance level \(\alpha\) (usually \(\alpha = 0.05\)). If \(p - value<\alpha\), we reject the null hypothesis.

Since \(p - value\approx0.0000<0.05\), the correct decision is to reject the null hypothesis.

Step 5: Conclusion

The correct summary would be: “There is sufficient evidence to conclude that the mean number of hours of all employees at start - up companies work more than the US mean of 47 hours.”

Final Answers (Filled in the Blanks)

• \(H_0\): \(\boldsymbol{\mu = 47}\)
• \(H_1\): \(\boldsymbol{\mu>47}\)
• Test Statistic: \(\boldsymbol{4.41}\) (example value, depends on actual data)
• p - value: \(\boldsymbol{0.0000}\) (example value, depends on actual data)
• Decision: Reject the null hypothesis
• Summary: “There is sufficient evidence to conclude that the mean number of hours of all employees at start - up companies work more than the US mean of 47 hours.”

(Note: The actual values for test statistic and p - value depend on the real sample data. The above is a demonstration with assumed values to show the process.)

Answer:

To solve this hypothesis testing problem (likely about the mean number of work hours), we follow these steps:

Step 1: Define the Hypotheses

The problem compares the mean work hours at start - up companies to the US mean of 47 hours.

• Null Hypothesis (\(H_0\)): This is the statement of no difference or the status quo. So, \(H_0: \mu = 47\) (the mean work hours at start - up companies is equal to 47 hours).
• Alternative Hypothesis (\(H_1\)): From the context “work more than the US mean”, we have a right - tailed test. So, \(H_1: \mu>47\) (the mean work hours at start - up companies is greater than 47 hours).

Step 2: Test Statistic (Assuming a z - test or t - test, let's assume we have sample data. For example, if we have a sample mean \(\bar{x}\), sample standard deviation \(s\), sample size \(n\))

The formula for the z - test statistic (when population standard deviation \(\sigma\) is known) is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). For a t - test (when \(\sigma\) is unknown), the formula is \(t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\).

Suppose we have a sample: Let's say \(n = 54\) (from the top - left box), sample mean \(\bar{x}\) (let's assume we calculated it as, for example, 50), population mean \(\mu = 47\), and sample standard deviation \(s = 5\) (these are assumed values for illustration, as the problem's full data is not shown).

For a t - test:
\(t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}=\frac{50 - 47}{\frac{5}{\sqrt{54}}}=\frac{3}{\frac{5}{7.3485}}\approx\frac{3}{0.6804}\approx4.41\) (rounded to 2 decimal places).

Step 3: p - value

For a right - tailed t - test with \(df=n - 1=54 - 1 = 53\) and test statistic \(t = 4.41\), we can use a t - distribution table or a calculator. Using a calculator (like the t - distribution function in statistical software), the p - value is very small, say approximately \(0.0000\) (rounded to 4 decimal places).

Step 4: Decision Rule

We compare the p - value to a significance level \(\alpha\) (usually \(\alpha = 0.05\)). If \(p - value<\alpha\), we reject the null hypothesis.

Since \(p - value\approx0.0000<0.05\), the correct decision is to reject the null hypothesis.

Step 5: Conclusion

The correct summary would be: “There is sufficient evidence to conclude that the mean number of hours of all employees at start - up companies work more than the US mean of 47 hours.”

Final Answers (Filled in the Blanks)

• \(H_0\): \(\boldsymbol{\mu = 47}\)
• \(H_1\): \(\boldsymbol{\mu>47}\)
• Test Statistic: \(\boldsymbol{4.41}\) (example value, depends on actual data)
• p - value: \(\boldsymbol{0.0000}\) (example value, depends on actual data)
• Decision: Reject the null hypothesis
• Summary: “There is sufficient evidence to conclude that the mean number of hours of all employees at start - up companies work more than the US mean of 47 hours.”

(Note: The actual values for test statistic and p - value depend on the real sample data. The above is a demonstration with assumed values to show the process.)