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STEP BY STEP Critical Value Approach to Hypothesis Testing 1- State H o and H 1 2- Choose level of significance, α Choose the sample size, n 3- Determine.

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Presentation on theme: "STEP BY STEP Critical Value Approach to Hypothesis Testing 1- State H o and H 1 2- Choose level of significance, α Choose the sample size, n 3- Determine."— Presentation transcript:

1 STEP BY STEP Critical Value Approach to Hypothesis Testing 1- State H o and H 1 2- Choose level of significance, α Choose the sample size, n 3- Determine the appropriate test statistics and sampling distribution. 4- Determine the critical values that divide the rejection and non-rejection areas. 5- Collect the sample data, organize the results and compute the value of the test statistics. 6- Make the statistical decision and state the managerial conclusion If the test statistics falls into non-rejection region, DO NOT REJECT H o If the test statistics falls into rejection region, REJECT H o The managerial conclusion is written in the context of the real world problem.

2 CORRECT DECISION 1-α TYPE II ERROR ( β ERROR) TYPE I ERROR (α ERROR) CORRECT DECISION 1-β DO NOT REJECT H 0 REJECT H 0 H 0 IS FALSE H 0 IS TRUE If the null hypothesis is true and accepted or false and rejected the decision is in either case CORRECT. If the null hypothesis is true and rejected or false and accepted the decision is in either case in ERROR. ACTUAL SITUATION STATISTICAL DECISION

3 Exercise - 1 One Tailed Test TEST at the 5% level whether the single sample value of 172 comes from a normal population with mean µ= 150 and variance σ 2 =100.

4 Exercise –McDonald One Tailed Test In one past study, McDonald’s had a mean service time of 174.22 seconds. Suppose that this company began a quality improvement effort to reduce the service time and selected a sample of 25 stores. The sample mean has been calculated as 162.96 seconds and sample standard deviation is 20.2 seconds. You wish to determine whether the new drive-through process has a mean that is less than 174.22 seconds.

5 Exercise –Internet Access Z-test for Proportion Of 2000 adults, 1540 said that they wanted Internet Access so, they could check personal e- mail while on vacation. A survey conducted in the previous year indicated that 75% of adults wanted Internet Access. Is there evidence that the percentage of adults who wanted Internet Access has changed from the previous year

6 Exercise – 3 The manager of the women`s dress department of a department store wants to know whether the true average number of women`s dresses sold per day is 24. If in a random sample of 36 days the average number of dresses sold is 23 with a standard deviation of 7 dresses, Is there, at the 0.05 level of significance, sufficient evidence to reject the null hypothesis that µ=24?

7 Exercise – 4

8 Exercise – 5

9 Exercise – Wind Speed A researcher claims that the average wind speed in a certain city 8 miles per hour. A sample of 32 days has an average wind speed of 8.2 miles per hour. The standard deviation of the sample is 0.6 mile per hour. At α = 0.05, is there enough evidence to reject the claim? Use p-value method.

10 Exercise – Starting Salary A job placement director claims that the average starting salary for nurses is 24,000 USD. A sample of 10 nurses` salaries has a mean of 23,450 USD and a standard deviation of 400 USD. Is there enough evidence to reject the director`s claim at α=0.05?

11 Exercise – 6

12 Exercise – Room Rates A survey found that the average hotel room rate in Antalya is 88.42 TL and the average room rate in Izmir 80.61 TL. Assume that the data were obtained from two samples of 50 hotels, each and that standard deviations were 5.63TL and 4.83 TL, respectively. At α = 0.05, can it be concluded that there is significant difference in the rates?


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