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This question was created from Tutorial Assignment 4 (CON 04 - Answers Q2 & Q3).docx https://www.coursehero.com/file/39624903/Tutorial-Assignment-4-CON-04-Answers-Q2-Q3docx/

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. A city health department wishes to determine if the mean bacteria count per unit volume of
water at a lake beach is below the safety level of 200. Bacteria levels in the lake are typically
normally distributed. A researcher collected 10 water samples of unit volume and found mean
and standard deviation of bacteria counts as 192.7 and 10.8, respectively.
(i) Is this a one-sided or two-sided test, why?
This is a one-sided test, because the interest is in bacteria levels below the safety limit.
(ii) Do the data indicate that there is no cause of concern (assume o=0.05)?
Hypotheses: null & alternative hypotheses:
Ho : # 2 200
HA : / < 200.
Test statistic: since population normal with unknown population variance, use t:
t = -
X - ul
s/ vn
= = to.05.9
Decision rule: n=10 and
0.05,9 =-1.833
reject null hypothesis if tots < -1.833
Observed test statistic: tabs = (192.7 - 200)/(10.8/ v10 ) =-2.14
Because -2.14 < -1.833, the data do not support the null hypothesis so we reject the null
hypothesis and conclude that there is no cause for concern (levels below safety limit)
(iii) Do the data indicate that there is no cause of concern if of=0.01?
At a 1% level of significance, the critical t = 2.821. Since -2.14 > -2.821, we do not have
sufficient evidence to reject the null, so we conclude that there is cause for concern
(levels above safety limit)
(iv) What is the p-value for the test statistic calculated in (ii), and what do you conclude
about the null hypothesis for a=0.05 & 0=0.01?
Can get p-value from online calculator (provided to students in lecture slides):
https://graphpad.com/quickcales/PValue1.cfm

Top Answer

(iv) p value = 0.0305 Reject null... View the full answer

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