An electronics company advertises that the battery life of its new smart phone with normal usage averages 72 hours. The company says this is one of the desirable characteristics of the phone which makes the phone a better choice compared to competitor phones. Recently, the company has received an unusually large number of complaints regarding the battery life. Many customers indicated the battery needed to be recharged long before the advertised average life. To check if the average battery life equals 72 hours, the company randomly selected 36 phones and tested the battery life. The sample produced a mean battery life of 69.5 hours with a standard deviation of 5 hours. Answer the following questions:
a. Perform the appropriate test to determine if the average battery life differs from the advertised life of 72 hours. Use the test statistic method to perform the test. Alpha for the test is .05. Be sure to include all required steps of the hypothesis test procedure as outlined and performed in class.
b. Compute a 95% confidence interval estimate for the mean battery life. How does your confidence interval support the conclusion of your hypothesis test?
a. To determine if the average battery life differs from the advertised life of 72 hours, we can perform a hypothesis test using the test statistic method.
Step 1: State the null and alternative hypotheses:
- Null Hypothesis (H0): The average battery life is equal to 72 hours.
- Alternative Hypothesis (Ha): The average battery life is different from 72 hours.
Step 2: Select the level of significance (Alpha):
- Alpha (α) is given as 0.05.
Step 3: Determine the test statistic and critical values:
- We'll use the t-test statistic because the population standard deviation is unknown and the sample size is less than 30.
Formula for the t-test statistic:
t = (sample mean - population mean) / (sample standard deviation / √sample size)
Given data:
Sample mean (x̄) = 69.5 hours
Population mean (µ) = 72 hours (advertised average)
Sample standard deviation (s) = 5 hours
Sample size (n) = 36
Step 4: Calculate the test statistic:
t = (69.5 - 72) / (5 / √36)
t = -2.5 / (5 / 6)
t = -2.5 / 0.8333
t ≈ -3
Step 5: Determine the critical region:
Since this is a two-tailed test, we need to compare the absolute value of the test statistic (-3) with the critical values from the t-distribution table.
At α = 0.05, with degrees of freedom (df) = 35 from a sample size of 36, the critical values are approximately ±2.030.
Step 6: Make a decision:
- If the absolute value of the test statistic is greater than the critical value (|-3| > 2.030), we reject the null hypothesis.
- If the absolute value of the test statistic is less than or equal to the critical value (|-3| ≤ 2.030), we fail to reject the null hypothesis.
Since |-3| > 2.030, we reject the null hypothesis.
b. To compute a 95% confidence interval estimate for the mean battery life, we use the following formula:
Confidence Interval Formula:
CI = x̄ ± (t * (s / √n))
Where:
CI = Confidence Interval
x̄ = Sample mean
t = Critical value from t-distribution table based on desired confidence level
s = Sample standard deviation
n = Sample size
Given data:
Sample mean (x̄) = 69.5 hours
Sample standard deviation (s) = 5 hours
Sample size (n) = 36
At 95% confidence level, the critical value from the t-distribution table with df = 35 is approximately ±2.030.
Plugging in the values, we have:
CI = 69.5 ± (2.030 * (5 / √36))
CI = 69.5 ± (2.030 * (5 / 6))
CI = 69.5 ± (2.030 * 0.8333)
CI ≈ 69.5 ± 1.6928
The confidence interval is approximately 69.5 ± 1.6928, which means the true mean battery life is estimated to fall between approximately 67.8072 hours and 71.1928 hours with 95% confidence.
The confidence interval supports the conclusion of the hypothesis test because the value of 72 hours (advertised average) is not within the range of the confidence interval.