Tutorial 08_1_2018_SOLUTIONS_18th April.pdf

Μ however since is unknown we have to use the

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𝑋 ̅ ~ 𝑁 , 𝜎 𝑛 ). However, since 𝜎 is unknown, we have to use the estimated std dev, S, hence 𝑋 ̅ − 𝜇 ? √𝑛 follows a t-distribution with n-1 degrees of freedom. Level of significance, α = .05 Claim μ < 970 (Must be H 1 since it does not have an ‘equals’ in it ).
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ETF1100 Business Statistics SOLUTIONS Tutorial 8 4 Step 1: Hypotheses H o : μ ≥ 970 H 1 : μ < 970 (one tailed test lower tail) Step 2: test statistic Population standard deviation unknown, use t distribution with 29 (30-1) degrees of freedom. ? = 𝑋 ̅ − µ 𝑜 ? √𝑛 t calc = 964−970 20 √30 = -1.643 Step 3: Critical Value with 29 (n-1: 30-1) degrees of freedom. tcrit ? .05,30−1 = -1.699 Step 4: Decision Rule and Decision Reject H 0 if t calc < -t crit Since -1.643 < -1.699, we cannot reject H o .
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ETF1100 Business Statistics SOLUTIONS Tutorial 8 5 Step 5: Conclusion We CANNOT reject H 0 at the 5% level of significance. The sample DOES NOT provide enough evidence against H 0 . That is, average parking revenue is NOT lower than $970. b) Explain the impact of your conclusion to the city council. I was provided with a sample of revenues for 30 weekdays. The mean revenue for these 30 days was indeed slightly less than $970; in fact it was $964. Even though the revenue is slightly lower than $970, we do expect some variation from sample to sample. So it would not be very surprising to get a result lower than $970, even if the consultant’s claim is true. However , based on the sample evidence, the mean revenue is not significantly lower than $9 70 so the consultant’s advice to the council was correct and their plan to pay for the building through parking fees seems to be on track. Q8.4 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
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