ADMS 2320 Final: Formula Sheet

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maroonelk473

22 Apr 2012

School

York University

Department

Administrative Studies

Course

ADMS 2320

Professor

Yogi

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Document Summary

We estimate with 95% confidence the mean weight of new products lies between. There is not enough evidence to infer that the standard deviation will not exceed. Sample size to estimate p , (if p is unknown, take . 5) 1%: only if a is more than 8% smaller than b ( v ) Conclusion: reject h0, there is sufficient evidence t. 0. 0022 < . 01( ) p-value< ,reject h0. Conclusion: reject h0, there is enough evidence to conclude. )( )()(222221212122221212221212/2122212121 nnsnnsnsnsnsnstxxnsnsxxt 2 n2 n1 v)(df 11)(2)1()1(11)()(2122/212122221122122121 nnstxxnnsnsnsnnsxxtppp 2221112/21n)p 1(p n)p 1(p z)p p (2 & 1 case for estimator 9704. 9652596. 1 estimate tosize 222/ wznsample ?)( or)/(/2/nstxnsxt )/(/2/nzxnxz 21,2/1221,2/2222)1()1()1( nnsnuclsnlclsn cc cnppzpnxpnppppz) 1( ,)1( 2/ 05. ,22,175,200 sxn180:180:10 hh2/2/,: zzorzzzzzzegionrejectionr 1,2/1,2/1,1,,: nnnnttorttttttegionrejectionr :isegionrejectionrthe:isstatistictestofvaluetheortt972. 1199,025. 2141. 3200/22180175 t972. 1199,025. tt)200/22(972. 1175 0677. 3175 21,2/221),2/(1221,221,12,: nnnnxxorxxxxxxegionrejectionr 2/2/,: zzorzzzzzzegionrejectionr 05. ,3800,20 sn2205000: h2215000: h1170. 10:219,95. 2 xxrr9744. 1050003800)120(222 c1,2/1,2/1,1,,: nnnnttorttttttegionrejectionr 4. 11014 ndxd7111. 2])([11222 nddnsd6465. 1 ds2/2/,: zzorzzzzzzrr 5. :5.

Related Questions

(1 point) Independent random samples, each containing 900 observations, were selected from two binomial populations. The samples from populations 1 and 2 produced 632 and 665 successes, respectively.

(a) Test against ≠ 0 . Use α = 0.07

Test statistic =

Rejection region lzl >

The final conclusion is

○ A. There is not sufficient evidence to reject the null hypothesis that

○ B. We can reject the null hypothesis that and accept that ≠ 0

(b) Test against . Use α = 0.02

Test statistic =

Rejection region z >

The final conclusion is

○ A. There is not sufficient evidence to reject the null hypothesis that

○ B. We can reject the null hypothesis that and accept that

oliver202166

1	Which of the following statements about Type I and Type II errors is correct
a	Type I: Reject a true alternative hypothesis. Type II: Do not reject a false alternative.
b	Type I: Reject a true null hypothesis. Type II: Do not reject a false null hypothesis.
c	Type I: Reject a false null hypothesis. Type II: Reject a true null hypothesis.
d	Type I: Do not reject a false null hypothesis. Type II: Reject a true null hypothesis.

2	You are reading a report that contains a hypothesis test you are interested in. The writer of the report writes that the p-value for the test you are interested in is 0.0749, but does not tell you the value of the test statistic. From this information you can:


a	Not reject the hypothesis at a Probability of Type I error = .05, but reject the hypothesis at a Probability of Type I error = 0.10

b	Reject the hypothesis at a Probability of Type I error = .05, and reject at a Probability of Type I error = 0.10

c	Not reject the hypothesis at a Probability of Type I error = 0.05, and not reject at a Probability of Type I error = 0.10

d	Reject the hypothesis at a Probability of Type I error = .05, but not reject at a Probability of Type I error = 0.10


3	The random sample below is obtained to test the following hypothesis about the population mean.
	H?: ? ?	1500
	H?: ? >	1500
	620	1711	366	2528	2678	1661	442	725	1938
	409	330	2480	542	369	2124	549	2074	1665
	1873	873	2143	2061	1177	2509	1264	2397	1523
	1837	1958	1041	1639	2199	2232	387	2270	2136
	1111	1883	2612	2230	1597	1726	694	1990	1354
	2090	909	2128	1608	747	1121	2220	2390	2347
	1041	316	655	632	2064	1901	532	552	846
	2704	1410	2165	1065	937	1452	2539	410	656
	1169	527	809	2364	2350	2210	1459	2391	856
	2711	1985	2382	2289	1927	518	2177	437	1151
	2018	1580	607	2715	2188	1691	1394	2610	1186
	695	2428	2246	858	2036	1681	2449	1578	1971
	1846	1729	2389	1737	1913	1863	2072	2593	2287
	2220	2230	551	458	2626	2731	488	2551	1736
	1373	307	1803	2647	2679	1508	1468	1443	516
	1002	2116	2616	817	2522	460	1879	1999	1837

	The level of significance of the test is ? = 0.05. Compute the relevant test statistic.
	This is a(n) _______ (two-tail, upper-tail, lower-tail) test. The test statistic is TS = _______.
a	Two-tail test		TS =	1.81
	Do not reject H?: ? ? 1500. Conclude that the population mean is not greater than 1500.
b	Upper tail test.		TS =	1.52
	Do not reject H?: ? ? 1500. Conclude that the population mean is not greater than 1500.
c	Upper tail test.		TS =	1.81
	Reject H?: ? ? 1500. Conclude that the population mean is greater than 1500.
d	Lower tail test.		TS =	1.98
	Do not reject H?: ? ? 1500. Conclude that the population mean is no greater than 1500.

4	Consider the following hypothesis test.
	H?: ? ?	30
	H?: ? >	30
	A random sample of n = 15 yielded the following observations
	51	38	26	16	28
	57	20	33	35	23
	21	47	56	54	36
	Use ? =	0.05
	TS = ______	CV = ______	State the decision rule.
a	1.68	1.761	Do not reject H?. Conclude the mean is not greater than 30.
b	1.68	1.64	Reject H?. Conclude the mean is greater than 30.
c	1.847	2.145	Do not reject H?. Conclude the mean is not less than 30.
d	1.847	1.761	Reject H?. Conclude the mean is less than 30.

5	In a recent study, a major fast food restaurant had a mean service time of 165 seconds. The company embarks on a quality improvement effort to reduce the service time and has developed improvements to the service process. The new process will be tested in a sample of stores. The new process will be adopted in all of its stores, if it reduced mean service time by more than 45 seconds compared to the current mean service time. To perform the hypothesis test, the sample of 48 stores yields the following data (seconds).





	90	96	133	108	136	110	119	138
	129	98	101	92	135	124	115	90
	132	125	110	124	126	138	94	130
	108	96	140	135	102	114	109	137
	138	104	108	134	92	107	96	119
	105	111	96	136	126	116	98	131

	Use ? =	0.05
	\|TS\| = ______	\|CV\| = ______
a	1.548	1.678	Do not reject H?. The mean service time is not reduced by more than 45 seconds. Do not adopt the new process.

b	1.871	1.678	Reject H?. The mean service time is reduced by more than 45 seconds. Adopt the new process.

c	1.871	1.640	Do not reject H?. The mean service time is not reduced by more than 45 seconds. Do not adopt the new process.

d	1.548	1.640	Reject H?. The mean service time is reduced by more than 45 seconds. Adopt the new process.

Please explain clearly and that pic/work is visible, thanks

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ADMS 2320 Final: Formula Sheet

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