Question

A man buys 60 electric bulbs from a company "P" and 70 bulbs from another company, "H". He finds that the average life of P's bulbs is 1500 hours with a standard deviation of 60 hours and the average life of H's bulbs is 1550 hours with a standard deviation of 70 hours. Then, the value of the test statistic to test that there is no significant difference between the mean lives of bulbs from the two companies, is:

• A. 2.85
• B. 4.38
• C. 5.27
• D. 3.90

Hint:

When sample sizes are large (typically n>30), the z-test is a robust choice for comparing means, even if the population standard deviations are unknown (we use sample standard deviations as estimates). Always check the sample sizes first to determine whether a z-test or a t-test is more appropriate.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The question asks for the value of the test statistic for a two-sample hypothesis test for the difference between two population means. The null hypothesis is that there is no difference between the mean lives of the bulbs from the two companies (\(H_0: \mu_P = \mu_H\)). Since the sample sizes are large (\(n_P = 60\) and \(n_H = 70\), both>30), we can use the z-test.

Step 2: Key Formula or Approach:
The formula for the two-sample z-test statistic is: \[ z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] Under the null hypothesis, \( \mu_1 - \mu_2 = 0 \). The sample standard deviations (\(s_1, s_2\)) are used to estimate the population standard deviations. 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: Detailed Explanation:
Let's 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 Now, substitute these values into the z-statistic formula: \[ z = \frac{1500 - 1550}{\sqrt{\frac{60^2}{60} + \frac{70^2}{70}}} \] \[ z = \frac{-50}{\sqrt{\frac{3600}{60} + \frac{4900}{70}}} \] \[ z = \frac{-50}{\sqrt{60 + 70}} \] \[ z = \frac{-50}{\sqrt{130}} \] Now, we calculate the value of \( \sqrt{130} \): \[ \sqrt{130} \approx 11.40175 \] \[ z = \frac{-50}{11.40175} \approx -4.3851 \] The value of the test statistic is the absolute value of z, which is approximately 4.385.
Step 4: Final Answer:
Rounding to two decimal places, the value of the test statistic is 4.38.