Question

Use the standard norml distribution or the t-distribution to construct a 99% confidence interval for the population mean. Justify your decision. If neither distribution. can be used,explain why. Interpret the results. In a random sample of 23 mortgage institutions,the mean interest rate was 3.65% and the standard deviation was 0.46%. Assume the interest rates are normally distributed.Which distribution should be used to construct the confidence interval?A. Use a normal distribution because the interest rates are normally distributed and is known.B. Use a t-distribution because it is a random sample, is unknown, and the interest rates are normally distributed.C. Use a normal distribution because n < 30 and the interest rates are normally distributed.D.Use a t-distribution because the interest rates are normally distributed is known.E. Cannot use the standard normal distribution or the t-distribution because is unknown, n <30, and the interest rates are not normally distributed.

Answer 1

Answer:

E) we will use t- distribution because is un-known,n<30

the confidence interval is (0.0338,0.0392)

Step-by-step explanation:

Step:-1

Given sample size is n = 23<30 mortgage institutions

The mean interest rate 'x' = 0.0365

The standard deviation 'S' = 0.0046

the degree of freedom = n-1 = 23-1=22

99% of confidence intervals $t_{0.01} =2.82$  (from tabulated value).

$The mean value = 0.0365$

$x±t_{0.01} \frac{S}{\sqrt{n-1} }$

$0.0365±2.82 \frac{0.0046}{\sqrt{23-1} }$

$0.0365±2.82 \frac{0.0046}{\sqrt{22} }$

$0.0365±2.82 \frac{0.0046}{4.690 }$

using calculator

$0.0365±0.00276$

Confidence interval is

$(0.0365-0.00276,0.0365+0.00276)$

$(0.0338,0.0392)$

the mean value is lies between in this confidence interval

(0.0338,0.0392).

Answer:-

using t- distribution because is unknown,n<30,and the interest rates are not normally distributed.