The mean life of 100 bulbs produced by the a company is
computed to be 1570 hours with standard deviation of 120 hours.
The company claims that the average life of bulbs produced by the
company is 1600 hours. using 5% level of significant , is the claim
acceptable?
To determine if the claim is acceptable at the 5% level of significance, we can perform a hypothesis test.
Null Hypothesis (H0): The average life of bulbs produced by the company is 1600 hours.
Alternative Hypothesis (H1): The average life of bulbs produced by the company is not 1600 hours.
We can set up the test statistic as follows:
t = (x̄ - μ) / (s / √n)
Where:
x̄ = sample mean (1570 hours)
μ = population mean (1600 hours)
s = standard deviation of the population (120 hours)
n = sample size (100 bulbs)
t = (1570 - 1600) / (120 / √100)
t = -30 / 12
t = -2.5
Now we can compare the calculated t-value with the critical t-value at 5% level of significance for a two-tailed test with 99 degrees of freedom. From a t-table, we find that the critical t-value is approximately ±2.626.
Since -2.5 falls outside of the range (-2.626, 2.626), we reject the null hypothesis.
Therefore, we have enough evidence to conclude that the claim made by the company (average life of bulbs produced by the company is 1600 hours) is not acceptable at the 5% level of significance.