a) The mean lifetime of a sample of 100 light tubes produced by a company is found to be 1570 hours with standard deviation of 80 hours. Test the hypothesis that the mean lifetime of the tubes produced by the company is 1600 hrs. (7 marks)
To test the hypothesis, we can perform a one-sample t-test. Let's go through the steps one by one:
Step 1: State the hypotheses.
- Null hypothesis (H₀): The mean lifetime of the tubes produced by the company is 1600 hours.
- Alternative hypothesis (H₁): The mean lifetime of the tubes produced by the company is not 1600 hours.
Step 2: Set the significance level (α).
Let's assume a significance level of 0.05, which is commonly used.
Step 3: Calculate the test statistic.
The test statistic for a one-sample t-test is given by:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
In this case:
Sample mean (x̄) = 1570 hours
Hypothesized mean (μ₀) = 1600 hours
Sample standard deviation (s) = 80 hours
Sample size (n) = 100
Let's substitute these values into the formula:
t = (1570 - 1600) / (80 / sqrt(100))
t = (-30) / (80 / 10)
t = -3.75
Step 4: Determine the critical value.
Since we are performing a two-tailed test, we need to find the critical values for a t-distribution with 99 degrees of freedom (n-1 = 100-1 = 99) and a significance level of 0.05.
Using a t-table or a statistical software, we find the critical values to be approximately tₐ/₂ = ±1.984.
Step 5: Make a decision.
If the absolute value of the test statistic (|-3.75|) is greater than the critical value (1.984), we reject the null hypothesis. If the test statistic is less extreme than the critical value, we fail to reject the null hypothesis.
In this case, |-3.75| > 1.984, so we reject the null hypothesis.
Step 6: Interpret the results.
Based on the sample data, there is sufficient evidence to conclude that the mean lifetime of the tubes produced by the company is not 1600 hours.
Step 7: Calculate the p-value (optional).
If you want to calculate the p-value, you can use a t-table or a statistical software. In this case, the p-value is less than the chosen significance level (0.05), which further supports rejecting the null hypothesis.
To test the hypothesis that the mean lifetime of the tubes produced by the company is 1600 hours, we can use a hypothesis testing method called a one-sample t-test. This test allows us to determine if there is enough evidence to support or reject the claim made in the hypothesis.
Here are the steps to perform the one-sample t-test:
Step 1: State the hypotheses:
The null hypothesis (H0): The mean lifetime of the tubes produced by the company is 1600 hrs.
The alternative hypothesis (Ha): The mean lifetime of the tubes produced by the company is not 1600 hrs.
Step 2: Set the significance level (α):
The significance level is the probability of rejecting the null hypothesis when it is true. In this case, let's set α = 0.05, which is a commonly used value.
Step 3: Calculate the test statistic:
The test statistic for a one-sample t-test is calculated using the formula:
t = (sample mean - hypothesized mean) / (standard deviation / square root of sample size)
Given:
Sample mean (x̄) = 1570 hrs
Hypothesized mean (µ) = 1600 hrs
Standard deviation (σ) = 80 hrs
Sample size (n) = 100
Plugging in the values, we get:
t = (1570 - 1600) / (80 / √100)
Step 4: Determine the critical value or p-value:
For a two-tailed t-test, we need to find the critical value or p-value associated with the calculated test statistic.
To find the critical value, we can consult a t-table with degrees of freedom (df) equal to (n - 1). In this case, df = 100 - 1 = 99. Looking up the critical value at α/2 = 0.025 and df = 99, we find it to be approximately ±1.984.
To find the p-value, we can use a statistical software or calculator. The p-value represents the probability of obtaining a test statistic as extreme as (or more extreme than) the observed value, assuming the null hypothesis is true.
Step 5: Make a decision:
- If the test statistic is outside the critical region or the p-value is less than the significance level (α), then we reject the null hypothesis.
- If the test statistic is inside the critical region or the p-value is greater than the significance level (α), then we fail to reject the null hypothesis.
In this case, calculate the test statistic value from Step 3 and compare it with the critical value or p-value from Step 4 to make a decision.
Step 6: Draw a conclusion:
Based on the decision made in Step 5, we can draw a conclusion about the hypothesis being tested.
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. Is that less than the level of significance you are using?