Questions: Suppose a test for a single mean, μ, is done. A sample of 36 observations is selected from a population that is N(μ, 9). The value of X̄=38.80. The test to be done is: H0: μ=38 H1: μ>38 The p-value for the test is: .1151 .0548 .1096

Suppose a test for a single mean, μ, is done. A sample of 36 observations is selected from a population that is N(μ, 9). The value of X̄=38.80. The test to be done is:
H0: μ=38 H1: μ>38

The p-value for the test is:
.1151
.0548
.1096

Transcript text: Suppose a test for a single mean, $\mu$, is done. A sample of 36 is observations is selected from a population that is $N(\mu, 9)$. The value of $\bar{X}=38.80$. The test to be done is: \[ H_{0}: \mu=38 \quad H_{1}: \mu>38 \] The $p$-value for the test is: .1151 .0548 .1096

Solution

Solution Steps

To find the p-value for the given hypothesis test, we need to calculate the test statistic using the sample mean, population mean, population standard deviation, and sample size. Then, we can use the test statistic to find the p-value from the standard normal distribution.

Step 1: Calculate the Test Statistic

Given:

Sample mean, $\bar{X} = 38.8$
Population mean, $\mu = 38$
Population standard deviation, $\sigma = 3$ (since the variance is 9, $\sigma = \sqrt{9} = 3$)
Sample size, $n = 36$

The test statistic (z-score) is calculated using the formula: \[ z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}} \]

Substituting the given values: \[ z = \frac{38.8 - 38}{\frac{3}{\sqrt{36}}} = \frac{0.8}{\frac{3}{6}} = \frac{0.8 \times 6}{3} = 1.6 \]

Step 2: Calculate the p-value

The p-value for a one-tailed test is calculated using the cumulative distribution function (CDF) of the standard normal distribution: \[ p\text{-value} = 1 - \Phi(z) \]

Where $\Phi(z)$ is the CDF of the standard normal distribution at $z = 1.6$.

Using the output: \[ p\text{-value} = 1 - \Phi(1.6) \approx 1 - 0.9452 = 0.0548 \]