Test Statistic & P-Value

Question

An article in the San Jose Mercury News stated that students in the California state university system take 6 years, on average, to finish their undergraduate degrees. A freshman student believes that the mean time is less and conducts a survey of 44 students. The student obtains a sample mean of 4.9 with a sample standard deviation of 0.7. Is there sufficient evidence to support the student's claim at an ∝ = 0.01 significance level?

Determine the test statistic. Round to four decimal places.

Find the p-value. Round to 4 decimals.

Answer 1

Hello Bailey, To determine the test statistic for this hypothesis test, you can perform a z-test for the population mean. The formula for the z-test statistic is:

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

Where:

- \(\bar{X}\) is the sample mean.

- \(\mu\) is the population mean under the null hypothesis (given as 6 years).

- \(\sigma\) is the population standard deviation (not provided, but you can use the sample standard deviation).

- \(n\) is the sample size (given as 44).

- \(\alpha\) is the significance level (given as 0.01).

Plug in the values:

\(\bar{X} = 4.9\) (sample mean)

\(\mu = 6\) (population mean under the null hypothesis)

\(\sigma = 0.7\) (sample standard deviation)

\(n = 44\) (sample size)

\(\alpha = 0.01\) (significance level)

Now, calculate the z-test statistic:

\[z = \frac{4.9 - 6}{\frac{0.7}{\sqrt{44}}}\]

Calculate the values in the denominator first:

\(\sqrt{44} \approx 6.6332\)

\(\frac{0.7}{6.6332} \approx 0.1054\)

Now, plug these values into the test statistic formula:

\[z = \frac{4.9 - 6}{0.1054} \approx -19.0286\]

Rounding to four decimal places, the test statistic is approximately -19.0286.

The next step would be to compare this test statistic with critical values from the standard normal distribution or find the p-value to determine whether there is sufficient evidence to reject the null hypothesis.