A machine put the juice in the bottle with normal distribution with mean is equal to 50 cl, with the standard deviation 5 cl. Test was done to see if the claim is right. A sample of 100 bottles were tested. They got the average content of 48.9 cl. At ∝ =0.05, can the claim be rejected?
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 the Z score.
To determine whether the claim can be rejected or not, we can conduct a hypothesis test using the given information. Here are the steps to conduct the hypothesis test:
Step 1: State the null and alternative hypotheses:
- Null hypothesis (H0): The mean content of juice in the bottles is 50 cl.
- Alternative hypothesis (Ha): The mean content of juice in the bottles is not 50 cl.
Step 2: Determine the significance level:
The significance level, denoted by α (alpha), is given as 0.05 in this case. It represents the maximum acceptable probability of rejecting the null hypothesis when it is true.
Step 3: Compute the test statistic:
To compute the test statistic, we can use the formula for the z-test for the population mean:
z = (x̄ - μ) / (σ / √n)
where:
- x̄ is the sample mean (48.9 cl in this case),
- μ is the population mean (50 cl),
- σ is the standard deviation of the population (5 cl), and
- n is the sample size (100 bottles).
Plugging in the values, we get:
z = (48.9 - 50) / (5 / √100) = -0.2
Step 4: Determine the critical value(s):
Since the alternative hypothesis is two-sided (not equal to), we need to find the critical value(s) from the standard normal distribution. With a significance level of α = 0.05, we split the significance level equally on both tails, giving us α/2 = 0.025 for each tail.
Looking up the critical value corresponding to a cumulative probability of 0.025 in the standard normal distribution table, we find that the critical value is approximately -1.96.
Step 5: Make a decision:
If the test statistic falls within the critical region (beyond the critical value(s)), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the test statistic z = -0.2 does not fall beyond the critical value of -1.96. Therefore, we fail to reject the null hypothesis.
Step 6: Interpret the result:
Since we failed to reject the null hypothesis, there is not enough evidence to conclude that the mean content of juice in the bottles is different from 50 cl. Thus, based on the given data and a significance level of α = 0.05, we cannot reject the claim that the machine puts the juice in the bottle with a mean of 50 cl.