Answer to Question #154655 in Statistics and Probability for fify

Question

Suppose an editor of a publishing company claims that the mean time to write a textbook is

less than 15 months. A sample of 16 textbook is randomly selected and it is found that the

mean time was 12.5 and SD was 3.6 months. Assuming the time to write a textbook is

normally distributed.

a. Construct a 90% confidence interval for μ, the mean time to write a textbook.

b. Using a 0.025 level of significance, would you conclude the editor’s claim is true?

Answer 1

Assuming a normal distribution ;

a) 90% CI=$\bar x±t_{\alpha/2, 15}. {\frac{s}{\sqrt n}}$

=$12.5±1.75×\frac{3.6}{\sqrt{16}}$

(10. 925, 14.075)

We are 90% confident that the population mean parameter lies between 10.925 and 14. 075.

b) we test the hypotheses;

$H_0:\mu=15$

$H_1:\mu<15$

$t=\frac{\bar x-\mu} {\frac {s} {\sqrt n}}$

$=\frac{12.5-15}{\frac{3.6}{\sqrt {16}}}$

=-2.78

The critical t value with 15 degrees of freedom is 2.131.

If the test statistic is less than the critical value we reject the null hypothesis.

-2.78<2.131

Since the test statistic is less than the critical value, we reject the null hypothesis in favor of the alternative.

We are 97.5% confident the mean time to write a textbook is less than 15 months.