Step 1: Concept Overview:
This problem requires calculating the test statistic for a two-sample hypothesis test comparing the means of two populations. The null hypothesis assumes no difference in the mean lifetimes of light bulbs from Company P and Company H (\(H_0: \mu_P = \mu_H\)). Given the large sample sizes (\(n_P = 60\) and \(n_H = 70\), both > 30), a z-test is appropriate.
Step 2: Formula:
The two-sample z-test statistic is calculated as:
\[ z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]
Under the null hypothesis, \( \mu_1 - \mu_2 = 0 \). We estimate population standard deviations using sample standard deviations (\(s_1, s_2\)).
Therefore, the formula simplifies to:
\[ z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]
Step 3: Calculation:
Define the parameters for each company:
Company P (Sample 1):
- Sample size: \( n_1 = 60 \)
- Sample mean: \( \bar{x}_1 = 1500 \) hours
- Sample standard deviation: \( s_1 = 60 \) hours
Company H (Sample 2):
- Sample size: \( n_2 = 70 \)
- Sample mean: \( \bar{x}_2 = 1550 \) hours
- Sample standard deviation: \( s_2 = 70 \) hours
Substitute these values into the z-statistic formula:
\[ z = \frac{1500 - 1550}{\sqrt{\frac{60^2}{60} + \frac{70^2}{70}}} \]
\[ z = \frac{-50}{\sqrt{\frac{3600}{60} + \frac{4900}{70}}} \]
\[ z = \frac{-50}{\sqrt{60 + 70}} \]
\[ z = \frac{-50}{\sqrt{130}} \]
Calculate \( \sqrt{130} \):\
\[ \sqrt{130} \approx 11.40175 \]
\[ z = \frac{-50}{11.40175} \approx -4.3851 \]
The test statistic is the absolute value of z, which is approximately 4.385.
Step 4: Answer:
Rounded to two decimal places, the test statistic is 4.38.