In a random sample of mobile devices, the mean repair cost was $ and the standard deviation was $. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a % confidence interval for the population mean. Interpret the results. The % confidence interval for the population mean is ( nothing, nothing). (Round to two decimal places as needed.)

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.

The margin of error is  

The 95% confidence interval  is

Step-by-step explanation:

From the question we are told that

  The sample mean is  

   The  standard deviation is  

   The sample size is  n = 5

Given the that the confidence level is  95% then the level  of significance is mathematically represented as

Next we obtain the critical value of   from the normal distribution table, the value is  

Generally the margin of error is mathematically represented as

substituting values

The 95% confidence interval  is mathematically represented as

substituting values

i got 86 for this

hope this !

if you need an explanation just ask

the equation for this situation is y=122+45. to solve the equation you would add 45 and 122 to get a total of 167lbs. in conclusion, grace's weight eight weeks ago was 167lbs.