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 one-sample t-test to evaluate the claim. The test statistic is calculated as:
t = (x̄ - μ) / (s / √n)
Where:
x̄ = sample mean (1570 hours)
μ = population mean (1600 hours)
s = sample standard deviation (120 hours)
n = number of bulbs (100 bulbs)
Plugging in the values, we get:
t = (1570 - 1600) / (120 / √100)
t = -30 / 12
t = -2.5
Using a t-distribution table or calculator, we find that the critical t-value at 5% significance level with 99 degrees of freedom is approximately ±2.626.
Since |-2.5| < 2.626, we fail to reject the null hypothesis. This means that there is not enough evidence to reject the company's claim that the average life of bulbs produced by the company is 1600 hours.
Themeanlife 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?
1 answer