The mean life of 100 bulbs produced by a company is computed 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 if the company's claim is acceptable, we can conduct 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 will use a t-test since we do not know the population standard deviation. The test statistic is given by:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Let's plug in the values:
Sample mean (x̄) = 1570 hours
Population mean (μ) = 1600 hours
Sample standard deviation (s) = 120 hours
Sample size (n) = 100 bulbs
Calculating the test statistic:
t = (1570 - 1600) / (120 / sqrt(100))
t = -30 / (120 / 10)
t = -2.5
Now, we need to determine the critical value for a two-tailed test at a 5% level of significance with (n-1) degrees of freedom. Since the sample size is 100, the degrees of freedom are 99.
Using a t-table or calculator, the critical value for a two-tailed test with 99 degrees of freedom at a 5% level of significance is approximately ±2.63.
Since -2.5 is not greater than ±2.63, we fail to reject the null hypothesis. This means there is not enough evidence to conclude that the average life of bulbs produced by the company is significantly different from 1600 hours at a 5% level of significance.
Therefore, the company's claim is acceptable at the 5% level of significance.