Closeness of a measurement to the actual value of the dimension being measured. For example, a measurement of 1.99999 cm for an
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Answer:
The 90% confidence interval for the true mean lifespan of this product is between 13.1 and 16.9 years.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 50 - 1 = 49
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 49 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.6766
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 15 - 1.9 = 13.1 years
The upper end of the interval is the sample mean added to M. So it is 15 + 1.9 = 16.9 years
The 90% confidence interval for the true mean lifespan of this product is between 13.1 and 16.9 years.