Complete Question
In a random sample of
five mobile devices, the mean repair cost was $75.00 and the standard deviation was $11.50
Assume the population is normally distributed and use at-distribution to find the margin of error and construct a 95%
confidence interval for the population mean. Interpret the results.
Answer:
The margin of error is [tex]E = 10.1[/tex]
The 95% confidence interval is [tex]64.9 < \mu < 85.1[/tex]
Step-by-step explanation:
From the question we are told that
The sample mean is [tex]\= x = \$ 75.00[/tex]
The standard deviation is [tex]\sigma = \$ 11.50[/tex]
The sample size is n = 5
Given the that the confidence level is 95% then the level of significance is mathematically represented as
[tex]\alpha = 100 -95[/tex]
[tex]\alpha = 5\%[/tex]
[tex]\alpha = 0.05[/tex]
Next we obtain the critical value of [tex]\frac{ \alpha }{2}[/tex] from the normal distribution table, the value is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{ \sqrt{n} }[/tex]
substituting values
[tex]E = 1.96* \frac{11.50 }{ \sqrt{5} }[/tex]
[tex]E = 10.1[/tex]
The 95% confidence interval is mathematically represented as
[tex]\= x - E < \mu < \= x + E[/tex]
substituting values
[tex]75 - 10.1< \mu < 75 + 10.1[/tex]
[tex]64.9 < \mu < 85.1[/tex]
