Medical researchers have developed a new artificial heart constructed primarily of titanium and plastic. The heart will last and operate almost indefinitely once it is implanted in the patient’s body, but the battery pack needs to be recharged about every four hours. A random sample of 50 battery packs is selected and subjected to a life test. The average life of these batteries is 4.05 hours. Assume that battery life is normally distributed with standard deviation o= 0.2 hour. Is there evidence to support the claim that mean battery life exceeds 4 hours? Use a= 0.05.
What is the test statistic for the test? Answer rounded-off to 2 decimal places
The Correct Answer and Explanation is :
To determine if there’s evidence that the mean battery life exceeds 4 hours, we perform a one-sample z-test for the population mean. Here’s how we proceed:
Given Data:
- Sample mean (( \bar{x} )) = 4.05 hours- Population mean (( \mu_0 )) = 4 hours- Population standard deviation (( \sigma )) = 0.2 hours- Sample size (( n )) = 50- Significance level (( \alpha )) = 0.05
Step 1: State the Hypotheses - Null hypothesis (( H_0 )): ( \mu = 4 ) hours (the mean battery life is 4 hours)- Alternative hypothesis (( H_1 )): ( \mu > 4 ) hours (the mean battery life exceeds 4 hours)
Step 2: Calculate the Test Statistic
The test statistic for a z-test is calculated using the formula:
[ z = \frac{\bar{x} – \mu_0}{\frac{\sigma}{\sqrt{n}}} ]
Substituting the given values:
[ z = \frac{4.05 – 4}{\frac{0.2}{\sqrt{50}} ]
Calculating the denominator:
[ \frac{0.2}{\sqrt{50}} \approx 0.028284 ]
Now, calculating the z-value:
[ z = \frac{0.05}{0.028284} \approx 1.77 ]
Step 3: Determine the Critical Value
For a one-tailed test at ( \alpha = 0.05 ), the critical z-value (( z_{\alpha} )) is approximately 1.645. This value corresponds to the point beyond which we would reject the null hypothesis.
Step 4: Make a Decision
- If the calculated z-value is greater than the critical z-value, we reject the null hypothesis.- If the calculated z-value is less than or equal to the critical z-value, we fail to reject the null hypothesis.
In this case, the calculated z-value is 1.77, which is greater than the critical z-value of 1.645. Therefore, we reject the null hypothesis.
Conclusion:
There is sufficient evidence at the 0.05 significance level to support the claim that the mean battery life exceeds 4 hours.