Q36SE Cortisol is a hormone that plays... [FREE SOLUTION]

91影视

Probability And Statistics For Engineering And Sciences

Jay L. Devore

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 714 pages

ISBN: 9781305251809

Chapter 10: Q36SE (page 435)

Cortisol is a hormone that plays an important role in mediating stress. There is growing awareness that exposure of outdoor workers to pollutants may impact cortisol levels. The article 鈥淧lasma Cortisol Concentration and Lifestyle in a Population of Outdoor Workers鈥� (Intl. J. of Envir. Health Res., 2011: 62鈥�71) reported on a study involving three groups of police officers: (1) traffic police (TP), (2) drivers (D), and (3) other duties (O). Here is summary data on cortisol concentration (ng/ml) for a subset of the officers who neither drank nor smoked.
Group
Sample Size
Mean
SD
TP
47
174.7
50.9
D
36
160.2
3702
O
50
153.5
45.9
Assuming that the standard assumptions for one-way ANOVA are satisfied, carry out a test at significance level .05 to decide whether true average cortisol concentration is different for the three groups. (Note: The investigators used more sophisticated statistical methodology (multiple regression) to assess the impact of age, length of employment, and drinking and smoking status on cortisol concentration; taking these factors into account, concentration appeared to be significantly higher in the TP group than in the other two groups.)

Short Answer

Expert verified

There is not sufficient evidence to support the claim that the true average cortisol concentration is different for the three groups.

Step by step solution

Finding value of ANOVA F

$\begin{aligned}{l}{n_1} = 47\\{n_2} = 36\\{n_3} = 50\\\overline {{x_1}} = 174.7\\\overline {{x_2}} = 160.2\\\overline {{x_3}} = 153.5\\{s_1} = 50.9\\{s_2} = 37.2\\{s_3} = 45.9\\\alpha = 0.05\end{aligned}$

The overall mean is the sum of all values divided by the number of values:

$\overline x = \frac{{{n_1}\overline {{x_1}} + {n_2}\overline {{x_2}} + {n_3}\overline {{x_3}} }}{N} = \frac{{47(174.7) + 36(160.2) + 50(153.5)}}{{47 + 3 + 50}} \approx 162.8053$

The mean square for treatment is:

$\begin{aligned}{l}MSTr = \frac{{{n_1}{{(\overline {{x_1}} - \overline x )}^2} + {n_2}{{(\overline {{x_2}} - \overline x )}^2} + {n_2}{{(\overline {{x_3}} - \overline x )}^2}}}{{I - 1}}\\ = \frac{{47{{(147.7 - 162.8053)}^2} + 36{{(160.2 - 162.8053)}^2} + 50{{(153.5 - 162.8053)}^2}}}{{3 - 1}} = 5611.7632\end{aligned}$

The mean square error is:

$\begin{aligned}{l}MSE = \frac{{({n_1} - 1)s_1^2 + ({n_2} - 1)s_2^2 + ({n_3} - 1)s_3^2}}{{N - I}}\\ = \frac{{(47 - 1){{(50.9)}^2} + (36 - 1){{(37.2)}^2} + (50 - 1){{(45.9)}^2}}}{{(47 + 36 + 50) - 3}} = 2083.4258\end{aligned}$

The ANOVA F statistic is the ratio of the MSTr and MSE:

$F = \frac{{MSTr}}{{MSE}} = \frac{{5611.7632}}{{2083.4258}} \approx 2.694$

Checking whether true average cortisol concentration is different for the three groups using P-Value

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of the F-distribution table in the appendix containing the F-value in the row$dfn = I - 1 = 3 - 1 = 2$and $dfd = N - I = 47 + 36 + 50 - 3 = 130 > 100:$

$0.050 < P < 0.100$

If the P-value is less than the significance level, then reject the null hypothesis.

$P > 0.05 \Rightarrow $Fail to reject${H_0}$

There is not sufficient evidence to support the claim that the true average cortisol concentration is different for the three groups.