A manufacturer claims that the mean lifetime of its lithium batteries is 1400 hours. A homeowner selects 25 of these batteries and finds the mean lifetime to be 1380 hours with a standard deviation of 80 hours. Test the manufacturers claim. Use alpha = 0.05. Round the test statistic to the nearest thousandth. Write the null and alternative hypotheses. Use the p-value method to vote for the null or the alternative hypothesis.
Additional related question.
If this same question states that the manufacture claim is greater than 1200 hours would the alternative hypothesis be ...
Ha: mean>1200
The null and alternative hypotheses for this test can be stated as follows:
Null hypothesis (H0): The mean lifetime of the lithium batteries is equal to 1400 hours.
Alternative hypothesis (H1): The mean lifetime of the lithium batteries is not equal to 1400 hours.
To test the manufacturer's claim, we will use a t-test since the population standard deviation is unknown. The test statistic can be calculated using the formula:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Given the following information:
Sample mean (x̄) = 1380 hours
Population mean (μ) = 1400 hours
Sample standard deviation (s) = 80 hours
Sample size (n) = 25
Using these values, we can calculate the t-test statistic as:
t = (1380 - 1400) / (80 / sqrt(25))
= (-20) / (80 / 5)
= (-20) / (16)
= -1.25 (rounded to the nearest thousandth)
Next, we need to find the critical t-value at a significance level of 0.05. Since this is a two-tailed test, we need to find the critical values in both tails.
With a sample size of 25 and a significance level of 0.05, the degrees of freedom for this test are (25 - 1) = 24.
To find the critical values, we can use a t-distribution table or a statistical software. For simplicity, let's assume the critical t-values are -2.064 and +2.064.
Comparing the t-test statistic (-1.25) with the critical values (-2.064 and +2.064), we can see that the t-test statistic falls within the range. Therefore, we fail to reject the null hypothesis.
To determine the p-value, we can use the t-distribution table or statistical software. For this two-tailed test with a t-test statistic of -1.25 and degrees of freedom of 24, the p-value is approximately 0.221.
Since the p-value (0.221) is greater than the significance level (0.05), we fail to reject the null hypothesis. This means that there is not enough evidence to support the claim that the mean lifetime of the lithium batteries is different from 1400 hours.
To test the manufacturer's claim, we will conduct a hypothesis test. First, we need to state the null and alternative hypotheses:
Null hypothesis (H0): The mean lifetime of the lithium batteries is equal to 1400 hours.
Alternative hypothesis (Ha): The mean lifetime of the lithium batteries is not equal to 1400 hours.
To test these hypotheses, we will use a t-test since the population standard deviation is unknown. Now, we can calculate the test statistic and p-value using the given sample statistics and significance level.
1. Calculate the test statistic (t-value):
The formula for the t-value is:
t = (sample mean - hypothesized mean) / (sample standard deviation / √n)
In this case:
Sample mean (x̄) = 1380 hours
Hypothesized mean (μ) = 1400 hours
Sample standard deviation (s) = 80 hours
Sample size (n) = 25
Plugging these values into the formula, we get:
t = (1380 - 1400) / (80 / √25)
= -20 / (80 / 5)
= -20 / 16
≈ -1.25
Round the test statistic to the nearest thousandth, we have t ≈ -1.250.
2. Calculate the p-value:
To find the p-value, we need to determine the probability of observing a t-value as extreme or more extreme than the one calculated, assuming the null hypothesis is true. Since the alternative hypothesis is two-tailed, we need to consider both tails.
Using a t-table or a t-distribution calculator, find the p-value associated with a t-value of -1.250 and degrees of freedom (df) equal to n - 1 (in this case, 25 - 1 = 24). The p-value is the probability of observing a t-value as extreme or more extreme than -1.250.
Let's assume the p-value is calculated as p = 0.225 (this is just an example for explanation purposes).
3. Compare the p-value to the significance level:
The significance level (alpha) is given as 0.05. If the p-value is less than or equal to alpha (p ≤ α), we reject the null hypothesis. If the p-value is greater than alpha (p > α), we fail to reject the null hypothesis.
In this case, p = 0.225, which is greater than 0.05 (p > α). Therefore, we fail to reject the null hypothesis.
In conclusion, based on the given data and using the p-value method, we do not have enough evidence to support the claim that the mean lifetime of the lithium batteries is different from 1400 hours.
Ho: mean = 1400
Ha: mean < 1400
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.