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Unformatted text preview: Find the critical valueﬂs} fer a lefttailed test for a population variance. sample size n = 23. and
level of signiﬁcance :1 = [LU25. The number of critical values depends on the type of hsptrthesis test. Since this is a lefttailed test there is one critical value. To determine the critical value ﬁrst
determine the degrees of ﬁ'eedtrm and the level of signiﬁcance. n  1 = 22 degrees of freedom
0‘. = H.025 You can either use a chisquare disuihutitrn table or techntrltrgv to ﬁnd the critical value. For this
explanation. a chisquare disuihulitrn table is used. This is a lefttailed test so the area in the left tail must he 0‘. = 0.025. Since the fdisuihulitrn tahle uses the area to the right of the values the critical value ctrrresptrnds ICI the area
1 — r1 = 0.9?5. New use a fdisuihulitrn table to ﬁnd the critical value uith 22 degrees of ﬁ'EEdDﬂl. I: a: 10.932 Find the critical valueIZs} for a twotailed test for a population variance. sample size n= 17". and
level of signiﬁcance or: ELIE. The number of critical values depends on the type of hvpothesis test. Since this is a twotailed test there are two critical values. To determine the critical values ﬁrst
determine the degrees of ﬁ'eedorn and the level of signiﬁcance. n 1 = 16 degrees of ﬁeedorn
ot=ELlll You can either use a chisquare dismhudon table or technology; to ﬁnd the critical value. For this
explanation. a chisquare disuihution table is used. 0" 'I . .
This is a twotailed test so the area in each tail must he F = ELDS. Since the fdism‘huuon table 0‘.
uses the area to the right of the values the critical values correspond to the areas F = [LES and l——=D.95.
2 Now use a fdism‘hution table to End the critical values with 16 degrees of ﬁ’eedorn. _I if =5 19:52 _I Xi =5 25.29:: State whether the standardized test statistic 3.52 allows you to
reject the null hypothesis. y? =25.3os Does the standardized test statistic 3:2 allow you to reject the null hypothesis? Reject the null hypothesis because the test statistic is in the rejection region. Use a jigtest to test the claim to {2.8 at the o = [L 1 El signiﬁcance level using sample statistics
s2 = 1.1 and n = 1?. Assume the population is normally distributed. 1iEl‘iil the null hypothesis he rejected at o = ﬂ. 1 El? Why or why not? 1_ 'I 'I
"‘r es_ hecause jg‘ 3’ IE.
_ 'I 'I
ho: because 3‘ :3 XE.
’ 'l 'l
Tr es_ because 35‘ s: E. No: because 3:2 s: Quarters are currentlv minted with weights normallv dismbuted and having a standard deviation of 0.069.
New equipment is being tested in an attempt to improve quality bf; reducing variation. A simple random
sample of 23 quarters is obtained from those manufactured with the new equipment and this sample has
a standard deviation of ELEMS. Use a [Hill signiﬁcance level to test the claim that quarters manufactured
with the new equipment have weights with a standard deviation less than 0.069. Does the new equipment
appear to be effective in reducing the variation of weights? (a) Write the claim mathematicallv and identify HE and Ha. A
The null and alternative hypotheses are HE: G 3: 0.069 and H1: 5 c: 0.069.
Eb) Find the critical valueEs). The number of critical values depends on the type of hypothesis test. HE: G 3: none and H1: 5 6. ELEM? is a lefttailed test. Since this is a lefttailed test there is one critical value. To determine the critical value ﬁrst determine
the degrees of freedom and the level of signiﬁcance. n 1 =2? degrees of ﬁeedom
:1: [Hill You can either use a chisquare dismbution table or technologv to ﬁnd the critical value. For this
explanation. a chisquare distnbution table is used. Ibis is a lefttailed test so the area in the left tail must be or = 0.01. Since the fdistnbuﬁon table uses
the area to the right of the values the critical value corresponds to the area 1  n = 0.99. Now use a fdismbution table to ﬁnd the critical value 'with 27‘ degrees of ﬁeedom. 33:512.?!9 Ihe rejection region for this test is shown to the right. 0: * I~J 12.3.79 I (c) Use the jailtest to ﬁnd the standardized test statistic. _, _ (n1)s2
I; _ 2
D"
_ [23—1)o_ou2
titular2
ss mass (d) Decide whether to reject or fail to reject the null hypothesis.
Compare the standardized test statistic to the critical value. The standardized test statistic is less than or equal to the critical value. Because this is a lefttailed test the null hspothesis will he rejected when 3:2 E j: [n this case. the null hypothesis can he rejected because the standardized test statistic 10.486 is less
than the critical value 12.8?9. (e) interpret the decision in the context of the original claim. It can he said that the nevi,r equipment appears to be effective because the null hypothesis is rejected. ﬂ Quarters are currendv minted with weights normallv dismhtned and having a standard devia1ion
of 0.061. New equipment is being tested in an attempt to improve quality hf; reducing variaIion.
A simple random sample of 29 quarters is obtained ﬁmn those manufactured with the new
equipment and this sample has a standard deviation of 0.048. Use a [LES signiﬁcance level to
test the claim that quarters manufactured with the new equipment have weights with a standard
devia1ion less than 0.061. Does the new equipment appear to he effective in reducing the
varia1ion of weights”?
(a) Write the claim mathemaIicallv and idenIifj; HE and Ha. A H31o E o_oo1; Ha: o} 0.061 (Claim) HE: G 2 one (Claim); Ha: o c: 0.061 HE: G E [12061 (Claim); Ha: o} 0.061 v‘ HE: G a o_oo1;Ha: o e oom (Claim)
Eh) Find the crilical valueEs).
353 = 16.928
(Use a comma to separate answers as needed. Round to three decimal places as needed} a
[den1ifv the rejec1ion regionﬂs).
if a: a: a: I: I: I:
(c) Use the xltest to ﬁnd the standardized test statis1ic. 352 = 1133? (Round to three decimal places as needed.) (d) Decide whether to reject or fail to reject the null hvpothesis. * Fail to reject I Reject (e) interpret the decision in the context of the original claim. Since the null hypothesis is rejected Since the null hypodlesis is rejected
the new equipment appears to he more the new equipment does not appear to
effective. he more effective. H" Since the null hypothesis is not Since the null hypothesis is not
rejected the new equipment rejected. the new equipment appears to does not appear to he more effective. he more effective. Suppose a mutual ﬁmd qualiﬁes as having moderate risk if the standard deviation of its monthly
rate of return is less than 3%. A mutualﬁmd rating agency randomly selects 30 months and
determines the rate of return for a certain ﬁrnd. The standard deviation of the rate of return is
computed to he 3.03 3%. Is there sufﬁcient evidence to support the claim that the ﬁmd has
moderate risk at the ct = [LBS level of signiﬁcance”? A normal prohahilr'ty plot indicates that the
monthly rates of return are normally dismhmed. (a) 1iErr"rite the claim mathernatic ally and identify HE and HE. A V" HUI! E 3; H315 6. 3 (Claim)
Hue: 3; Hanoi 3 (Claim) HEZD' E 3 (Claim); Hana 3’ 3
(lo) Find the critical valueﬂs}. Kg = 11mg
(Use a comma to separate answers as needed. Round to three decimal places as needed.) Idemify the rejection region[s).
u"
a: a: a:
I: I: I:
(c) Use the jigtest to End the standardized test statistic. 352 = 29.383 (Round to three decimal places as needed.) (d) Daaida whathar ta rajaat or fail to rajaat tha null hypothasis
H" Fall to raj aat Raj act (a) Intarprat tha daaisinn in tha anntaxt nftha miginal Claim. Is thara sufﬁaiant avidanaa to support tha claim that tl1a mutual ﬁJnd has nmdarata risk at
tha :1: [LBS las‘al of sigtu'ﬁaanaa? V I‘M—u: baaausa tha standardizad tast statistic is not in tha rajaatiun raginn. Nu: baaausa tha standardizad tast statistic is in tha rajaatitrn I‘Egitrﬂ. V Raj aat Fall to rajaat (a) [ntarprat tha daaisinn in tha auntaxt uftha original Claim. Sinaa tha null hypothasis is not Sinaa tha null hypnthasis is rajaatai raj aatai tha naw aqtﬁpmant appaars to tha naw aquipmant duas nut appaar to
ba mnra affaativa. ha nmra affaativa. Sinaa tha null h}?ﬂtl1&5i5 is not H“ Sinaa tha null hypnthasis 1's rajaataui raj aatai tha naw aqtﬁprnant tha naw aqm'pmant appaars to ba nmra daas not appaar to ha mnra affaativa. affaativa. ...
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This note was uploaded on 12/30/2009 for the course MATH 1410 taught by Professor Vergo during the Spring '09 term at Metropolitan Community College Omaha.
 Spring '09
 vergo
 Statistics

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