Question

The R. R. Bowker Company collects information on the retail prices of books and publishes the data in The Bowker Annual Library and Book Trade Almanac. In 2005, the mean retail price of agriculture books was $57.61. This year’s average retail price for n = 28 randomly selected agriculture books was ¯x = $54.97 with standard deviation of s = $8.45. (a) Use significance level ↵ = 0.05 and conduct the following test H0 : µ = 57.61 versus Ha : µ 6= 56.71 (b) Interpret your answer in part (a).

Answer 1

Answer:

a) $t=\frac{54.97-57.61}{\frac{8.45}{\sqrt{28}}}=-1.653$    

The degrees of freedom are given by:

$df=n-1=28-1=27$  

$p_v =2*P(t_{(27)}$  

b) The conclusion depends on the significance level selected, if the p value is lower than a significance level given we have enough evidence to conclude that the true mean is different from 57.61, otherwise we FAIl to reject the null hypothesis and thare is no evidence to conclude that the true mean is different from 57.61

Step-by-step explanation:

Data provided

$\bar X=54.97$ represent the mean for the retail price

$s=8.45$ represent the sample standard deviation

$n=28$ sample size selected  

$\mu_o =57.61$ represent the value that we want to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value for the test

Part a

System of hypothesis

We want to check if the true mean for the reatil price is 57.61 or not, the system of hypothesis would be:  

Null hypothesis:$\mu = 57.61$  

Alternative hypothesis:$\mu \neq 57.61$  

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

The statistic is given by:

$t=\frac{54.97-57.61}{\frac{8.45}{\sqrt{28}}}=-1.653$    

P value

The degrees of freedom are given by:

$df=n-1=28-1=27$  

The p value would be:

$p_v =2*P(t_{(27)}$  

Part b: Conclusion  

The conclusion depends on the significance level selected, if the p value is lower than a significance level given we have enough evidence to conclude that the true mean is different from 57.61, otherwise we FAIl to reject the null hypothesis and thare is no evidence to conclude that the true mean is different from 57.61