The mean lifetime of a sample of 100 light tubes is 1570 hours with a standard deviation of 80 hours test the hypothesis that the mean lifetime of the tubes is 1600 hours

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 of the Z score.

To test the hypothesis that the mean lifetime of the tubes is 1600 hours, you can perform a hypothesis test using the sample mean and the given information.

Here are the steps to test this hypothesis:

Step 1: State the null and alternative hypotheses:
- Null hypothesis (H0): The mean lifetime of the tubes is equal to 1600 hours.
- Alternative hypothesis (H1): The mean lifetime of the tubes is not equal to 1600 hours.

Step 2: Set the significance level (α):
- Choose a significance level (α) to determine the cutoff value for the test. Commonly used values are 0.05 or 0.01. For this example, let's use α = 0.05.

Step 3: Calculate the test statistic:
- Use the formula for the test statistic:
t = (sample mean - hypothesized mean) / (standard deviation / √(sample size))

For this example:
Sample mean (x̄) = 1570 hours
Hypothesized mean (μ0) = 1600 hours
Standard deviation (σ) = 80 hours
Sample size (n) = 100

Plugging in the values:
t = (1570 - 1600) / (80 / √100) = -30 / 8 = -3.75

Step 4: Determine the critical region and critical value:
- Since we have a two-tailed test (we are testing whether the mean is not equal to 1600 hours), we need to find the critical values in both tails.
- Look up the critical value for a t-distribution with (n - 1) degrees of freedom and the chosen significance level (α). For our example, with (100 - 1) = 99 degrees of freedom and α = 0.05, the critical value is approximately ±1.984.

Step 5: Make a decision:
- Determine whether the test statistic falls in the critical region or not.
- If the absolute value of the test statistic is greater than the critical value, reject the null hypothesis; otherwise, fail to reject the null hypothesis.

In this case, |t| = |(-3.75)| = 3.75 > 1.984 (critical value). Therefore, we reject the null hypothesis.

Step 6: State the conclusion:
- Based on the data and the hypothesis test, there is enough evidence to conclude that the mean lifetime of the tubes is not equal to 1600 hours, with a significant level of 0.05.

To test the hypothesis that the mean lifetime of the light tubes is 1600 hours, we can use a hypothesis testing approach. In this case, we will use a one-sample t-test.

Here are the steps to perform the hypothesis test:

Step 1: Formulate the null and alternative hypotheses
- Null hypothesis (H0): The mean lifetime of the light tubes is equal to 1600 hours.
- Alternative hypothesis (Ha): The mean lifetime of the light tubes is not equal to 1600 hours.

Step 2: Define the significance level (α)
- The significance level, denoted as α, determines the threshold for accepting or rejecting the null hypothesis. Commonly used values for α include 0.05 or 0.01.

Step 3: Compute the test statistic
- The test statistic for a one-sample t-test is calculated as:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the hypothesized population mean (1600 hours), s is the sample standard deviation, and n is the sample size.

Step 4: Determine the critical value
- The critical value(s) are obtained from the t-distribution table based on the degrees of freedom and the chosen significance level (α).

Step 5: Compare the test statistic with the critical value
- If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Step 6: Draw conclusions
- Based on the outcome of the hypothesis test, we can draw conclusions about whether there is sufficient evidence to support the alternative hypothesis.

In this case, we are given the sample mean (x̄ = 1570 hours), the sample standard deviation (s = 80 hours), and the sample size (n = 100).

Let's calculate the test statistic and check if we can reject the null hypothesis:

t = (1570 - 1600) / (80 / √100)
t = -30 / 8
t = -3.75

Next, we need to determine the critical value for our chosen significance level (α), which is not provided in the question. Let's assume α = 0.05 (5% significance level), which is commonly used.

Looking up the critical value for a two-tailed test at α = 0.05 and degrees of freedom (df = n - 1 = 99) in the t-distribution table, we find that the critical value is approximately ±1.984.

Since the absolute value of the test statistic (-3.75) is greater than the critical value (1.984), we can reject the null hypothesis.

Conclusion: There is sufficient evidence to suggest that the mean lifetime of the light tubes is different from 1600 hours based on the given sample.