Answer to Question #199823 in Statistics and Probability for Aljohn R jaen

Question

The mean scores of a random sample of 17 students who took a special test is 83.5. If the standard deviation of the scores is 4.1, and the sample comes from an approximately normal population, find the interval estimates of the population mean adopting a confidence level of 90%.

Task:

1.The point estimate;

2.The interval estimate adopting the 95%confidence level;

3.determine if a certain student passed the special test if it's score is 80; and

4.conclusion and recommendations with regards to the special test.

Answer 1

Mean, $\bar{x}=83.5$

Sample size, $n=17$

Standard deviation, $\sigma=4.1$

Z at 90% confidence level = 1.645

$s_M=\sqrt{\dfrac{(4.1)^2}{17}}=0.99$

Interval estimate,

$\mu=\bar x\pm (Z\times s_M)$

$\mu=83.5\pm (1.645\times0.99)$

$\mu=83.5\pm1.62$

(1) Point estimate = Sample mean = 83.5

(2) Z at 95% confidence level = 1.96

$\mu=\bar x\pm (Z\times s_M)$

$\mu=83.5\pm (1.96\times0.99)$

$\mu=83.5\pm1.949$

(3) Since, 80 is less than the lower bound of interval estimate calculated above i.e

80 does not lie in the interval [81.551, 85.449].

Therefore, we can say that the student did not pass the special test.

(4) Conclusion :

Such special tests tell us about the progress of students with regard to their studies on a daily basis.

Recommendation :

The frequency of such tests should be increased as to increase the overall mean of the class