The Acme Company has developed a new battery. The engineer in charge claims that the new battery will operate continuously for at leas 7 minutes longer than the old battery. To test the claim, the company selects a simple random sample of 100 new batteries and 100 old batteries. The old batteries run continuously for 190 minutes with a standard deviation of 20 minutes; the new batteries, 200 minutes with a standard deviation of 40 minutes. Test the engineer's claim that the new batteries run at least 7 minutes longer than the old. Use a 0.05 level of significance. (assume that therre are no outliers in either sample.)
Use the appropriate statistical formula for this type of 2-sample test. Calculate the test statistic using the information in the problem. Find your critical value using the appropriate table at .05 level of significance for a one-tailed test. Compare your test statistic to the critical value from the table. If the test statistic exceeds the critical value from the table, reject the null. If the test statistic does not exceed the critical value, do not reject the null. Draw your conclusions from the results.
To test the engineer's claim, we can perform a hypothesis test using the sample data provided.
Step 1: State the Hypotheses
Let's assume:
H0: The mean operating time of the new batteries is less than or equal to the mean operating time of the old batteries.
Ha: The mean operating time of the new batteries is greater than the mean operating time of the old batteries.
Step 2: Determine the Level of Significance
The level of significance, denoted by alpha, is given as 0.05.
Step 3: Calculate the Test Statistic
Since we are comparing two means, we can use the two-sample t-test formula to calculate the test statistic:
t = (x̄1 - x̄2 - d0) / sqrt((s1^2 / n1) + (s2^2 / n2))
Where:
x̄1 and x̄2 are the sample means of the old and new batteries respectively.
s1 and s2 are the sample standard deviations of the old and new batteries respectively.
n1 and n2 are the sample sizes of the old and new batteries respectively.
d0 is the hypothesized difference in means (which is 7 in this case).
Step 4: Determine the Critical Value
Since we are conducting a one-tailed test (to the right), we need to find the critical value for the given level of significance (0.05) and degrees of freedom (n1 + n2 - 2).
Step 5: Calculate the P-value
Using the t-distribution and the test statistic calculated in Step 3, we can find the corresponding p-value.
Step 6: Make a Decision
If the p-value is less than the level of significance (0.05), we reject the null hypothesis (H0). Otherwise, we fail to reject the null hypothesis.
Note: In this step, we are testing if the p-value is less than 0.05 since we are making a claim that the new batteries run at least 7 minutes longer.
Step 7: State the Conclusion
Based on the decision made in Step 6, we conclude whether there is enough evidence to support the engineer's claim.
Now, let's calculate the test statistic and perform the hypothesis test using the provided data.
x̄1 = 190 (mean operating time of the old batteries)
s1 = 20 (standard deviation of the old batteries)
n1 = 100 (sample size of the old batteries)
x̄2 = 200 (mean operating time of the new batteries)
s2 = 40 (standard deviation of the new batteries)
n2 = 100 (sample size of the new batteries)
d0 = 7 (hypothesized difference in means)
Step 3: Calculate the Test Statistic
t = (x̄1 - x̄2 - d0) / sqrt((s1^2 / n1) + (s2^2 / n2))
= (190 - 200 - 7) / sqrt((20^2 / 100) + (40^2 / 100))
= -17 / sqrt(4 + 16)
= -17 / sqrt(20)
≈ -3.80
Step 4: Determine the Critical Value
For a one-tailed test with a level of significance of 0.05 and (n1 + n2 - 2) degrees of freedom (198 in this case), the critical value is approximately 1.653.
Step 5: Calculate the P-value
Using the t-distribution with (n1 + n2 - 2) degrees of freedom and the test statistic (-3.80), we can find the p-value associated with this test.
Step 6: Make a Decision
Since the p-value is less than the level of significance (0.05), we reject the null hypothesis (H0).
Step 7: State the Conclusion
Based on the decision made in Step 6, there is enough evidence to support the engineer's claim that the new batteries run at least 7 minutes longer than the old batteries.
To test the engineer's claim that the new batteries run at least 7 minutes longer than the old batteries, we can perform a hypothesis test.
Step 1: Define the null and alternative hypotheses.
- Null hypothesis (H0): The new batteries do not run at least 7 minutes longer than the old batteries. The mean difference in battery operation time (μnew - μold) is less than or equal to 7 minutes.
- Alternative hypothesis (H1): The new batteries run at least 7 minutes longer than the old batteries. The mean difference in battery operation time (μnew - μold) is greater than 7 minutes.
Step 2: Select a significance level.
- The given significance level is 0.05, which means we are willing to accept a 5% chance of committing a Type I error (rejecting the null hypothesis when it is true).
Step 3: Compute the test statistic.
- We can use the formula for the test statistic for comparing two independent sample means with unequal variances:
t = (x̄new - x̄old - D) / sqrt((snew^2 / nnew) + (sold^2 / nold))
where x̄new and x̄old are the sample means, snew and sold are the sample standard deviations, nnew and nold are the sample sizes, and D is the claimed difference (7 minutes).
Substituting the given values:
x̄new = 200 minutes, x̄old = 190 minutes, snew = 40 minutes, sold = 20 minutes, nnew = 100, nold = 100, and D = 7 minutes.
t = (200 - 190 - 7) / sqrt((40^2 / 100) + (20^2 / 100))
= 3 / sqrt(16 + 4)
= 3 / sqrt(20)
≈ 0.67
Step 4: Determine the critical value(s).
- Since the alternative hypothesis is one-sided (the new batteries run at least 7 minutes longer), we need to compare the test statistic to the critical value of the t-distribution at the given significance level and degrees of freedom (df = nnew + nold - 2).
- Using a significance level of 0.05 and the degrees of freedom of 198 (100 + 100 - 2), the critical t-value is approximately 1.653 (obtained from a t-table or calculator).
Step 5: Make a decision.
- If the test statistic is greater than the critical value, we reject the null hypothesis.
- If the test statistic is less than or equal to the critical value, we fail to reject the null hypothesis.
The test statistic (0.67) is less than the critical value (1.653). Therefore, we fail to reject the null hypothesis.
Step 6: State the conclusion.
- Based on the given data and the hypothesis test, there is not enough evidence to support the engineer's claim that the new batteries run at least 7 minutes longer than the old batteries.