Question

a) construct a 95% confidence interval using the Z or Interval b) state which interval was used - Z or T Interval c) interpret your answer in context of the problem In a random sample of 50 people , the mean body mass index (BMI) was 27.7 and the sample standard deviation was 6.12

Answer 1

a) Use the following expression for the confidence interval:

`\bar{x}\pm Z_{(\alpha)/(2)}\frac{\sigma}{\sqrt[]{n}}`

where the first term is the mean of the body mass index of the sample (27.7), sigma is the standard deviation (6.12) and n is the number of the sample (50); alpha is the confidence level, given by:

α = 1- 0.95 = 0.05

α/2 = 0.05/2 = 0.025

The value of the normal distribution Z for 0.025 is, by searching in a table, 1.96:

Z_0.025 = 1.96

replace the values of the parameters into the expression to determine the confidence interval:

C.I = {27.7 - (1.96)(6.12/√50) , 27.7 + (1.96)(6.12/√50)}

C.I = {26.01 , 29.39}

b) Due to the sample is greater than 30, the distribution used is the normal distribution Z.

c) The result obtained means that with a 95% of confidence, you can conclude that values around the mean are 26.01 and 29.39 for the body mass indexof the sample. That is, you can be sure that the mean body mass index of the sample is in between 26.01 and 29.39, with a 95% confidence.