Section 7.5B - Find the critical valueﬂs} fer a...

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Unformatted text preview: Find the critical valueﬂs} fer a left-tailed test for a population variance. sample size n = 23. and level of signiﬁcance :1 = [LU-25. The number of critical values depends on the type of hs-ptrthesis test. Since this is a left-tailed 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 chi-square disuihutitrn table or techntrltrgv to ﬁnd the critical value. For this explanation. a chi-square disuih-ulitrn table is used. This is a left-tailed test so the area in the left tail must he 0‘. = 0.025. Since the f-disuihulitrn 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 f-disuihulitrn table to ﬁnd the critical value uith 22 degrees of ﬁ'EEdDﬂl. I: a: 10.932 Find the critical valueIZs} for a two-tailed 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 two-tailed 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 chi-square dismhudon table or technology; to ﬁnd the critical value. For this explanation. a chi-square disuihution table is used. 0" 'I . . This is a two-tailed test so the area in each tail must he F = ELDS. Since the f-dism‘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 f-dism‘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 jig-test 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 left-tailed test. Since this is a left-tailed 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 chi-square dismbution table or technologv to ﬁnd the critical value. For this explanation. a chi-square distnbution table is used. Ibis is a left-tailed test so the area in the left tail must be or = 0.01. Since the f-distnbuﬁon table uses the area to the right of the values the critical value corresponds to the area 1 - n = 0.99. Now use a f-dismbution 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 jail-test to ﬁnd the standardized test statistic. _, _ (n-1)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 left-tailed 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 xl-test 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 mutual-ﬁmd 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 jig-test 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.

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