You wish to test the following claim $$ \left(H_{a}\right) $$ at a significance level of $$ \alpha=0.001 $$. $$ H_{o}: \mu=77.9 $$ $$ \quad H_{a}: \mu

Question

You wish to test the following claim $$ \left(H_{a}\right) $$ at a significance level of $$ \alpha=0.001 $$. $$ H_{o}: \mu=77.9 $$ $$ \quad H_{a}: \mu<77.9 $$ You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size $$ n=30 $$ with mean $$ M=71.7 $$ and a standard deviation of $$ S D=10.3 $$. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) $$ \alpha $$ greater than $$ \alpha $$ This test statistic leads to a decision to... reject the null accept the null fail to reject the null

UpStudy AI · Accepted Answer

To conduct the hypothesis test, we will first calculate the test statistic using the formula for the t-test since the population standard deviation is unknown.

### Step 1: Calculate the test statistic

The formula for the t-test statistic is:

$$
t = \frac{M - \mu_0}{\frac{S}{\sqrt{n}}}
$$

Where:
- $$ M $$ is the sample mean
- $$ \mu_0 $$ is the population mean under the null hypothesis
- $$ S $$ is the sample standard deviation
- $$ n $$ is the sample size

Given:
- $$ M = 71.7 $$
- $$ \mu_0 = 77.9 $$
- $$ S = 10.3 $$
- $$ n = 30 $$

Substituting the values into the formula:

$$
t = \frac{71.7 - 77.9}{\frac{10.3}{\sqrt{30}}}
$$

Calculating the denominator:

$$
\frac{10.3}{\sqrt{30}} \approx \frac{10.3}{5.477} \approx 1.875
$$

Now substituting back into the t formula:

$$
t = \frac{71.7 - 77.9}{1.875} = \frac{-6.2}{1.875} \approx -3.293
$$

So, the test statistic is:

$$
\text{test statistic} \approx -3.293
$$

### Step 2: Calculate the p-value

To find the p-value, we need to look up the t-distribution with $$ n - 1 = 30 - 1 = 29 $$ degrees of freedom for the calculated t-value of -3.293.

Using a t-table or calculator, we find the p-value corresponding to $$ t = -3.293 $$ with 29 degrees of freedom.

Using statistical software or a calculator, we find:

$$
\text{p-value} \approx 0.001
$$

### Step 3: Compare the p-value to the significance level

Given that $$ \alpha = 0.001 $$:

- The p-value $$ \approx 0.001 $$ is **less than or equal to** $$ \alpha $$.

### Step 4: Decision

Since the p-value is less than or equal to $$ \alpha $$, we reject the null hypothesis.

### Summary of Results

- Test statistic: $$ \text{test statistic} = -3.293 $$
- p-value: $$ \text{p-value} = 0.0010 $$
- The p-value is less than (or equal to) $$ \alpha $$.
- This test statistic leads to a decision to reject the null.
- test statistic: -3.293 	- p-value: 0.0010 	- The p-value is less than (or equal to) $$ \alpha $$. 	- This test statistic leads to a decision to reject the null.

Upstudy AI Solution

Answer

Solution

Step 1: Calculate the test statistic

Step 2: Calculate the p-value

Step 3: Compare the p-value to the significance level

Step 4: Decision

Summary of Results

Beyond the Answer

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