Question

For the following problem : a ) construct a 95 % confidence interval using the Z or T Interval b ) state which interval was used - Z or T Interval c ) interpret your answer in context of the problem Assume the data for all the problems are normally distributed .In a random sample of 15 mortgage institutions , the mean interest rate was 3.57% and the sample standard deviation was 0.36 %.

Answer 1

In a random sample of 15 mortgage institutions.

Sample size = n = 15

The mean interest rate was 3.57% and the sample standard deviation was 0.36%

Sample mean = x = 3.57%

Sample standard deviation = s = 0.36%

a) construct a 95% confidence interval using the Z or T Interval

The 95% confidence interval is given by

`CI=\bar{x}\pm t_{(\alpha)/(2)}\cdot\frac{s}{\sqrt[]{n}}`

Where tα/2 is the t-value, from the t-table, the t-value at 95% confidence level and 14 degree of freedom (15 - 1 = 14)

tα/2 = 2.145

`\begin{gathered} CI=\bar{x}\pm t_{(\alpha)/(2)}\cdot\frac{s}{\sqrt[]{n}} \\ CI=3.57\pm2.145\cdot\frac{0.36}{\sqrt[]{15}} \\ CI=3.57\pm2.145\cdot0.0929 \\ CI=3.57\pm0.199 \\ CI=3.57-0.199,\: 3.57+0.199 \\ CI=(3.37,\: 3.77) \end{gathered}`

Therefore, the 95% confidence interval using t-interval is (3.37%, 3.77%)

b) state which interval was used - Z or T Interval

Since the population standard deviation was unknown, and the population was normally distributed hence t-interval was used.

c) interpret your answer in the context of the problem

The 95% confidence interval (3.37%, 3.77%) means that we are 95% confident that the mean interest rate will lie between this interval (3.37%, 3.77%)

It can be as low as 3.37% or as high as 3.77% with 95% confidence