Answer to Question #335746 in Statistics and Probability for anjoann

Question

Student scores on a test that measures self-image are approximately normally distributed. This test is

administered to 20 science students and the mean and standard deviation of their test scores are 88

and 24, respectively.

a) Find a 98 % confidence interval of the true mean.

b) What can we assert with 98% confidence about the possible size of our error if we estimate the

mean score to be 88?

Answer 1

The critical value for $\alpha = 0.02$ and $df = n-1 = 19$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.539483.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}},\bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$ $$=(88-2.539483\times\dfrac{24}{\sqrt{20}},$$

$$88+2.539483\times\dfrac{24}{\sqrt{20}})$$

$$=(74.3717,101.6283)$$

2. Based on the data provided, the 98% confidence interval for the population mean is $74.3717 < \mu < 101.6283,$ which indicates that we are 98% 

confident that the true population mean $\mu$ is contained by the interval $(74.3717, 101.6283).$

$$SE=\dfrac{s}{\sqrt{n}}=\dfrac{24}{\sqrt{20}}\approx5.3666$$