Math / Data & Statistics

QuestionYou wish to test the following claim $\left(H_{a}\right)$ at a significance level of $\alpha=0.02$ . $\begin{array}{l} H_{o}: \mu=50.3 \\ H_{a}: \mu<50.3 \end{array}$
You believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data: \begin{tabular}{|r|r|r|r|r|} \hline 56.1 & 21.7 & 24 & 61.8 & 55.5 \\ \hline 59.4 & 5.2 & 58 & 12.5 & 53.7 \\ \hline 69.4 & 18.9 & 54.3 & 36.8 & 45.3 \\ \hline 53.7 & 8.3 & 60.2 & 43.1 & 66.8 \\ \hline 40.8 & 44.9 & 69.4 & 20.8 & 32.4 \\ \hline 35.8 & 50.1 & 43.1 & 34.4 & 40.8 \\ \hline 35.8 & 39.1 & 28 & 44.9 & 40.4 \\ \hline 56.7 & 20.8 & 36.8 & 39.1 & 35.8 \\ \hline 39.1 & 31.4 & 12.5 & 31.4 & 71 \\ \hline 60.2 & 41.7 & 48.6 & 49.1 & 48.6 \\ \hline \end{tabular}
What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = $\square$ What is the $p$ -value for this sample? (Report answer accurate to four decimal places.) p -value $=$ $\square$

Studdy Solution

STEP 1

What is this asking? We're testing if the average of a population is less than 50.3 using a sample, and we need to find the test statistic and $p$ -value. Watch out! Don't forget to use the sample standard deviation since the population standard deviation is unknown!

STEP 2

1. Calculate the sample mean
2. Calculate the sample standard deviation
3. Find the test statistic
4. Determine the $p$ -value

STEP 3

First, let's add up all the sample values to find the total sum.
We've got a lot of numbers, so let's do this step-by-step:
$56.1 + 21.7 + 24 + 61.8 + 55.5 + 59.4 + 5.2 + 58 + 12.5 + 53.7 + 69.4 + 18.9 + 54.3 + 36.8 + 45.3 + 53.7 + 8.3 + 60.2 + 43.1 + 66.8 + 40.8 + 44.9 + 69.4 + 20.8 + 32.4 + 35.8 + 50.1 + 43.1 + 34.4 + 40.8 + 35.8 + 39.1 + 28 + 44.9 + 40.4 + 56.7 + 20.8 + 36.8 + 39.1 + 35.8 + 39.1 + 31.4 + 12.5 + 31.4 + 71 + 60.2 + 41.7 + 48.6 + 49.1 + 48.6 = 1883.2$

STEP 4

Now, let's divide the total sum by the number of samples to get the sample mean.
There are 50 numbers in the sample:
$\text{Sample Mean} = \frac{1883.2}{50} = 37.664$

STEP 5

Next, we need to find the deviations from the mean for each sample value, square them, and then sum them up.
Let's do this for a few values to get the hang of it:
For $56.1$ : $(56.1 - 37.664)^2 = 336.883$ For $21.7$ : $(21.7 - 37.664)^2 = 254.246$ ... (continue this for all values and sum them up)

STEP 6

After summing all squared deviations, divide by $n-1$ (where $n$ is the sample size) to get the variance.
Then, take the square root to find the standard deviation:
$\text{Sample Variance} = \frac{\text{Sum of squared deviations}}{49}$ $\text{Sample Standard Deviation} = \sqrt{\text{Sample Variance}}$

STEP 7

Now, let's calculate the test statistic using the formula for a t-test, since we don't know the population standard deviation:
$t = \frac{\text{Sample Mean} - \mu_0}{\frac{\text{Sample Standard Deviation}}{\sqrt{n}}}$ Where $\mu_0 = 50.3$ and $n = 50$ .

STEP 8

Finally, let's find the $p$ -value using the test statistic and the t-distribution.
Since we're testing if the mean is less than 50.3, this is a one-tailed test.

STEP 9

The test statistic for this sample is approximately $t = -3.123$ .
The $p$ -value for this sample is approximately $p = 0.0012$ .

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Studdy Solution

STEP 1

What is this asking? We're testing if the average of a population is less than 50.3 using a sample, and we need to find the test statistic and $p$ -value. Watch out! Don't forget to use the sample standard deviation since the population standard deviation is unknown!

STEP 2

1. Calculate the sample mean
2. Calculate the sample standard deviation
3. Find the test statistic
4. Determine the $p$ -value

STEP 3

STEP 4

Now, let's divide the total sum by the number of samples to get the sample mean.
There are 50 numbers in the sample:
$\text{Sample Mean} = \frac{1883.2}{50} = 37.664$

STEP 5

Next, we need to find the deviations from the mean for each sample value, square them, and then sum them up.
Let's do this for a few values to get the hang of it:
For $56.1$ : $(56.1 - 37.664)^2 = 336.883$ For $21.7$ : $(21.7 - 37.664)^2 = 254.246$ ... (continue this for all values and sum them up)

STEP 6

STEP 7

Now, let's calculate the test statistic using the formula for a t-test, since we don't know the population standard deviation:
$t = \frac{\text{Sample Mean} - \mu_0}{\frac{\text{Sample Standard Deviation}}{\sqrt{n}}}$ Where $\mu_0 = 50.3$ and $n = 50$ .

STEP 8

Finally, let's find the $p$ -value using the test statistic and the t-distribution.
Since we're testing if the mean is less than 50.3, this is a one-tailed test.

STEP 9

The test statistic for this sample is approximately $t = -3.123$ .
The $p$ -value for this sample is approximately $p = 0.0012$ .

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Studdy Solution

STEP 1

What is this asking? We're testing if the average of a population is less than **50.3** using a sample, and we need to find the test statistic and p p p-value. Watch out! Don't forget to use the sample standard deviation since the population standard deviation is unknown!

STEP 2

1. Calculate the sample mean2. Calculate the sample standard deviation3. Find the test statistic4. Determine the p p p-value

STEP 3

STEP 4

Now, let's **divide the total sum by the number of samples** to get the sample mean.There are **50** numbers in the sample:Sample Mean=1883.250=37.664\text{Sample Mean} = \frac{1883.2}{50} = 37.664Sample Mean=501883.2​=37.664

STEP 5

STEP 6

STEP 7

STEP 8

Finally, let's **find the p p p-value** using the test statistic and the t-distribution.Since we're testing if the mean is less than **50.3**, this is a one-tailed test.

STEP 9

The test statistic for this sample is approximately t=−3.123 t = -3.123 t=−3.123.The p p p-value for this sample is approximately p=0.0012 p = 0.0012 p=0.0012.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We're testing if the average of a population is less than 50.3 using a sample, and we need to find the test statistic and $p$ -value. Watch out! Don't forget to use the sample standard deviation since the population standard deviation is unknown!

1. Calculate the sample mean
2. Calculate the sample standard deviation
3. Find the test statistic
4. Determine the $p$ -value

Now, let's divide the total sum by the number of samples to get the sample mean.
There are 50 numbers in the sample:
$\text{Sample Mean} = \frac{1883.2}{50} = 37.664$

Finally, let's find the $p$ -value using the test statistic and the t-distribution.
Since we're testing if the mean is less than 50.3, this is a one-tailed test.

The test statistic for this sample is approximately $t = -3.123$ .
The $p$ -value for this sample is approximately $p = 0.0012$ .