The mean life of 100 bulbs produced by a company is assumed
to be 1570 hours with standard deviation of 120hours. the
company claims that the average life of bulbs produced by the
company is 1600 hours. using 5% level of significance, is the
claim acceptable?
To determine whether the claim that the average life of bulbs produced by the company is 1600 hours is acceptable, we can conduct a hypothesis test.
Let's state our null and alternative hypotheses:
Null hypothesis (H0): The average life of bulbs produced by the company is equal to 1600 hours.
Alternative hypothesis (H1): The average life of bulbs produced by the company is not equal to 1600 hours.
We can conduct a two-tailed t-test since we are not given any specific direction in the alternative hypothesis.
The test statistic can be calculated using the formula:
t = (x̄ - μ) / (s/√n)
Where:
x̄ is the sample mean (1570 hours),
μ is the claimed average (1600 hours),
s is the sample standard deviation (120 hours),
and n is the sample size (100 bulbs).
Plugging in the values, we have:
t = (1570 - 1600) / (120/√100)
= -30 / (120/10)
= -30 / 12
= -2.5
Now, we need to determine the critical t-value at a 5% level of significance with 99 degrees of freedom (n - 1 = 100 - 1 = 99). Looking up this value in a t-table or using a statistical software, we find the critical t-value to be approximately ±1.984.
Since the absolute value of our calculated t-value (-2.5) is greater than the critical t-value (1.984), we can reject the null hypothesis.
Therefore, we have sufficient evidence to conclude that the claim that the average life of bulbs produced by the company is 1600 hours is not acceptable at the 5% level of significance.