Question

A school librarian purchases a novel for her library. The publisher claims that the book is written at a 5th grade reading level, but the librarian suspects that the reading level is higher than that. The librarian selects a random sample of 40 pages and uses a standard readability test to assess the reading level of each page. The mean reading level of these pages is 5.2 with a standard deviation of 0.8. Do these data give convincing evidence at the = 0.05 significance level that the average reading level of this novel is greater than 5?

Answer 1

`t=(5.2-5)/((0.8)/(√(40)))=1.58`

The degrees of freedom are given by:

`df=n-1=40-1=39`

Thep value for this case would be given by:

`p_v =P(t_((39))>1.58)=0.061`

Since the p value is higher than the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that true mean is not significantly higher than 5.

Explanation:

Information provided

`\bar X=5.2` represent the sample mean

`s=0.8` represent the sample standard deviation

`n=40` sample size

`\mu_o =5` represent the value to verify

`\alpha=0.05` represent the significance level

t would represent the statistic

`p_v` represent the p value

Hypothesis to test

We want to verify if the true mean is higher than 5, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 5`

Alternative hypothesis:
`\mu > 5`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info given we got:

`t=(5.2-5)/((0.8)/(√(40)))=1.58`

The degrees of freedom are given by:

`df=n-1=40-1=39`

Thep value for this case would be given by:

`p_v =P(t_((39))>1.58)=0.061`

Since the p value is higher than the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that true mean is not significantly higher than 5.