# Tutorial 9 Solutions S1 2015

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Tutorial 9 Part A A9.1 In hypothesis testing, the level of significance (α) is a very important concept. Explain what this means.

α is the probability of rejecting the null hypothesis when it is true. In other words, we decide the risk level we are willing to have in rejecting the null when it is true. (refer to Citiprop eg in Lecture 8 - there is only 5% probability that Z will take a value larger than 1.645; if in fact CitiProp is selling houses in 42 days or less).

α is usually set at 0.01, 0.05 or 0.10 (also referred to as percentage eg, 1%, 5% or 10%).

A9.2 State the 6 “Helpful Hints” for hypothesis testing (Week 8 Lecture notes). 1) State the Claim first, determine if it is H0 or H1,then write the complementary hypothesis. 2) Null hypothesis H0 must ALWAYS have an equals sign in it. So, it must be one of: “=“

“≤”

“≥”

3) H0 and H1 must be complementary. 4) H1 tells us whether it is a one or a two – tailed test. 5) Use value of parameter assumed to be true under H0 in Zcalc / tcalc formulae (eg μ0). 6) Make decision in terms of H0, but interpret conclusion with reference to H1 (because this is what the data tells you about).

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A9.3 For the following textbook questions, state the null and alternative hypotheses (H0 and H1) ONLY. You are NOT REQUIRED to complete the rest of the question, just state H0 and H1. Berenson, page 284 onwards: 9.16, 9.18, 9.19, 9.43, 9.44, 9.45. 9.16 Ho : µ ≤10 H1 : µ >10

9.18 Ho : µ = 4 H1 : µ ≠ 4

9.19 Ho : µ = 10 H1 : µ ≠ 10

9.43 Ho : µ ≥ 855 H1 : µ < 855

9.44 Ho : µ ≥ 25 H1 : µ < 25 9.45 Ho : µ ≤ 10 H1 : µ > 10 2

A9.4 A Chinese factory produces mother-of-pearl buttons. The mean diameter of buttons is assumed to be normally distributed. If the machinery is working correctly, the mean diameter of the buttons will be 1.3cm. The variance for that machine is known to be 0.0081 cm 2. A sample of 16 buttons is measured with a laser, and found to have a mean diameter of 1.25cm. Test at the 5% level of significance, the hypothesis that the mean diameter of the population differs from 1.3cm. Ensure that you clearly state your hypotheses, show ALL steps, ALL your working AND interpret your conclusion in context of this question.

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Step 5: Conclusion We CAN reject H0 at the 5% level of significance. The sample DOES provide enough evidence against H0. Therefore the mean diameter is significantly different to 1.3cm. The machine is not working properly.

A9.5 Eighteen per cent of multinational companies provide an allowance for personal long-distance calls for executives living overseas. Suppose a researcher thinks that multinational companies are having a more difficult time recruiting executives to live overseas and that an increasing number of companies are providing an allowance for personal long-distance calls to these executives to ease the burden of living away from home. To test this claim, a new study is conducted by randomly sampling 376 multinational companies. Twenty two per cent of these surveyed companies are providing an allowance for personal long-distance calls to executives living overseas. Using this information and the 1% level of significance, test this claim. Ensure that you clearly state your hypotheses, show ALL steps, ALL your working AND interpret your conclusion in context of this question. Let x= multinational companies that provide an allowance for personal longdistance calls for executives living overseas n =376; p = 0.22; α = 0.01 nπ=376 x 0.18=67.68 and n(1-π)=376(1-0.18)=308.32 Note that both are greater than 5, hence the normal approximation is valid. Claim: π > 0.18 4

Step 1: Hypotheses

H 0 :   0.18 H1 :   0.18

one-tailed test. Therefore, Zα = Z0.01 = 2.327

Step 2:Test statistic

zcalc 

zcalc 

p  is distributed approximately as N(0,1)  (1   ) n 0.22  0.18  2.02 0.18(1  0.18) 376

Step 3:Critical Value Critical value: zcrit  z  z0.01  2.327

Reject H0 Z 2.327

Step 4:Decision Reject the null hypothesis if zcalc  zcrit Since 2.02 > 2.327 we CANNOT reject the null hypothesis.

Step 5: Conclusion

We CANNOT reject H0 at the 1% level of significance. The sample DOES NOT provide enough evidence against H0. That is the proportion of multinational companies provide an allowance for personal long-distance calls for executives living overseas in NOT significantly higher (has not increased). 5

Part B B9.6 Question 9.28(a), page 291 of textbook The director of manufacturing at a fabric mill needs to determine whether a new machine is producing a particular type of cloth according to the manufacturer’s specifications, which indicate that the cloth should have a mean breaking strength of 30kg and a standard deviation of 3.5kg. A sample of 49 pieces of cloth reveals a sample mean breaking strength of 29.3kg. Is there evidence that the machine is not meeting the manufacturer’s specifications for mean breaking strength? Use a 5% level of significance. Ensure that you clearly state your hypotheses, show ALL steps, ALL your working AND interpret your conclusion in context of this question.

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Step 5: Conclusion We CANNOT reject H0 at the 5% level of significance. The sample DOES NOT provide enough evidence against H0. Therefore the mean breaking strength is NOT significantly different to 30kg. The new machine is producing cloth according to the manufacturer’s specifications.

B9.7 Hoping to lure more shoppers downtown, a regional city council builds a new public parking garage close to the main shopping street. The city plans to pay for the building through parking fees. The consultant who advised the city on this project predicted that parking revenues would average \$970 a day. For a random sample of 30 weekdays, the daily fees collected averaged \$964. a) Assuming the population standard deviation is \$20, perform a hypothesis test to determine whether there is evidence at the 5% level of significance that the revenue is lower than the consultant predicted. Ensure that you clearly state your hypotheses, show ALL steps, ALL your working AND interpret your conclusion in context of this question.

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Step 5: Conclusion We CANNOT reject H0 at the 5% level of significance. The sample DOES NOT provide enough evidence against H0. Therefore parking revenue is NOT significantly lower than \$970.

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b) Explain the impact of your conclusion to the city council.

B9.8 A study by a research consultant showed that 79% of companies offer flexible scheduling. Suppose a researcher believes that in accounting companies this figure is lower. The researcher randomly selects 415 accounting companies, and through interviews determines that 303 of these companies have flexible scheduling. Using a 1% level of significance, test if there is enough evidence to conclude that a significantly lower proportion of accounting companies offer employees flexible scheduling. Ensure that you clearly state your hypotheses, show ALL steps, ALL your working AND interpret your conclusion in context of this question. Let X=companies that offer flexible scheduling. n =415;

p=

303 = 0.7301; 415

α = 0.01

nπ=415 x 0.79=327.85and n(1-π)=415(1-0.79)=87.15 Note that both are greater than 5, hence the normal approximation is valid. Claim: π < 0.79 Step 1: Hypotheses

H 0 :   0.79 H1 :   0.79

one-tailed test(left tailed test). Therefore, Zα = Z0.01 =-2.327

Step 2:Test statistic

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zcalc 

p  0.7301  0.79   3  (1   ) 0.79(1  0.79) n 415

Step 3:Critical Value Critical value: zcrit  z  z0.01  2.327

Reject H0

Z -2.327

Step 4:Decision Reject the null hypothesis if zcalc   zcrit Since -3 < -2.327, we CAN reject the null hypothesis. Step 5: Conclusion We CAN reject H0 at the 1% level of significance. The sample DOES provide enough evidence to against H0. That is the proportion of accounting companies that offer employees flexible scheduling is significantly lower than 79%

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