Complete Question
In a random sample of ten people, the mean driving distance to work was 23.1 miles and the standard deviation was 6.6 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 99% confidence interval for the population mean Interpret the results. Identify the margin of error.
Answer:
The 99% confidence interval is [tex]16.32< \mu <29.88[/tex]
The interpretation is that there is 99% confidence that the true mean lies within the limits
The margin of error is [tex]E = 6.783[/tex]
Step-by-step explanation:
From the question we are told that
The sample mean is [tex]\= x = 23.1[/tex]
The standard deviation is [tex]\sigma = 6.6 \ miles[/tex]
The sample size is n = 10
Generally the degree of freedom is mathematically represented as
[tex]df = n-1[/tex]
=> [tex]df = 10-1[/tex]
=> [tex]df =9[/tex]
Given that the confidence level is 99% , the n the level of significance is mathematically evaluated as
[tex]\alpha = 100 - 99[/tex]
[tex]\alpha =1\%[/tex]
[tex]\alpha =0.01[/tex]
Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] with a df of 9 from from the student t-distribution table the value is
[tex]t _{\frac{\alpha }{2} , df } = 3.250[/tex]
Generally the margin of error is mathematically represented as
[tex]E = t_{\frac{\alpha }{2} , df } * \frac{\sigma }{\sqrt{n} }[/tex]
[tex]E = 3.250 * \frac{6.6 }{\sqrt{10} }[/tex]
[tex]E = 6.783[/tex]
The 99% confidence interval is
[tex]\= x - E < \mu < \= x + E[/tex]
=> [tex]23.1 - 6.78 < \mu <23.1 + 6.78[/tex]
=> [tex]16.32< \mu <29.88[/tex]
The interpretation is that there is 99% confidence that the true mean lies within the limits
