Exercise :

A machine is set up such that the average content of juice per bottle equals μ . A sample of 36 bottles yeilds an average content of 48.5cl. Test the hypothesis that the average content per bottle is 50cl at the significance level.

Assume that the population standard deviation σ = 5cl

Indicate your specific subject in the "School Subject" box, so those with expertise in the area will respond to the question.

What level of significance are you using? P = .05? P = .01?

Z = (score-mean)/SEm

SEm = SD/√n

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to your Z score. Is it higher or lower than your significance level?

To test the hypothesis that the average content per bottle is 50cl at the given significance level, we will perform a hypothesis test using the z-test.

Step 1: State the hypotheses.
The null hypothesis (H0) states that the average content per bottle is 50cl.
H0: µ = 50

The alternative hypothesis (HA) states that the average content per bottle is not equal to 50cl.
HA: µ ≠ 50

Step 2: Set the significance level.
The significance level (α) is the threshold for determining statistical significance. Let's assume that the significance level is 0.05, which means we want to be 95% confident in our results.

Step 3: Calculate the test statistic.
Since we know the population standard deviation (σ) and the sample size (n), we can use the z-test formula:

z = (x̄ - µ) / (σ / √n)

where x̄ is the sample mean, µ is the population mean, σ is the population standard deviation, and n is the sample size.

Plugging in the values, we get:

z = (48.5 - 50) / (5 / √36)
z = -1.5 / (5 / 6)
z = -1.5 / (0.83333)
z ≈ -1.8

Step 4: Determine the critical value(s).
Since we are performing a two-tailed test (≠), we need to find the critical values that correspond to the significance level of 0.05/2 = 0.025. Looking up the z-value in the standard normal distribution table, the critical z-values are approximately -1.96 and 1.96.

Step 5: Make a decision.
If the test statistic falls within the critical values range, we fail to reject the null hypothesis. If the test statistic falls outside the critical values range, we reject the null hypothesis.

In this case, the test statistic z ≈ -1.8 falls in the rejection region (z < -1.96 for a two-tailed test), so we reject the null hypothesis.

Step 6: Interpret the result.
Based on the sample data, with a significance level of 0.05, we have enough evidence to conclude that the average content per bottle is not equal to 50cl. The average content is significantly different from 50cl.

Note: It's important to note that this is just one example of how to approach the hypothesis test. Different textbooks or statistical software may use slightly different formulas or approaches.

To test the hypothesis that the average content per bottle is 50cl, we can use a one-sample t-test. This test compares the sample mean to a known or hypothesized population mean when the population standard deviation is unknown.

Here are the steps to perform the t-test:

Step 1: State the null and alternative hypotheses
Null hypothesis (H0): The average content per bottle equals 50cl (μ = 50)
Alternative hypothesis (H1): The average content per bottle does not equal 50cl (μ ≠ 50)

Step 2: Determine the significance level (α)
The significance level, often denoted as α, is the probability of rejecting the null hypothesis when it is true. In this case, the significance level is not mentioned, so let's assume it to be 0.05 (5%).

Step 3: Set up the t-test
The t-test statistic is given by the formula:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))

In this case, the sample mean is 48.5cl, the hypothesized mean is 50cl, the sample standard deviation is 5cl, and the sample size is 36.

t = (48.5 - 50) / (5 / sqrt(36))
t = -1.5 / (5 / 6)
t = -1.5 / 0.8333
t ≈ -1.8

Step 4: Determine the critical value(s)
The critical value(s) depends on the significance level (α) and the degrees of freedom (df), which is (sample size - 1). For a two-tailed test at a 5% significance level, the critical value for a sample size of 36 can be found using a t-table or statistical software. Let's assume the critical value is approximately ±2.0301.

Step 5: Make a decision and interpret the results
If the t-test statistic falls within the critical value range, we fail to reject the null hypothesis. If the t-test statistic falls outside the critical value range, we reject the null hypothesis.

In this case, -1.8 does not exceed the critical value of ±2.0301. Therefore, we fail to reject the null hypothesis. We do not have enough evidence to conclude that the average content per bottle is significantly different from 50cl.

Note: Performing a t-test assumes that the sample is randomly selected and is normally distributed. If these assumptions are not met, alternative tests may be required.