QUESTION 26 If we decide the probability of making at Type 1 error is too great

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QUESTION 26 If we decide the probability of making at Type 1 error is too great at the .05 level of significance for our ?rooms and wo decide to use a 01 level of significance will our decision and conclusion change; if so why? The revised problem would then be: A statewide real estate company specializes in selling farm property in the state of Nebraska. Their records ndicate that the mean selling time of farm property is 90 days. Because of recent drought conditions, they believe that the mean selling time is now greater than 90 days. A statewide survey of 101 farms sold recently revealed that the mean selling time was 94 days, with a standard deviation of 22 days. At the 01 significance level you will test to determine if there has been an increase in selling time. What will be our decision and conclusion? T T ! ? Paragraph : Arial : 3(12pn :-. 1: . T . , . . Path: p Wordso QUESTION 27 The policy of the Suburban Transit authority is to add a bus route it moee than 55 percent of the potential commuters indicate they would use the particular route. A sample of 70 commuters revealed that 42 would use a proposed route from Bowman Park to the downtown area. You will be testing to determine if the Bowman-to-downtown route meets the STA criterion using the 05 significance level. What is your Alternate hypothesis (HA T T T ? Paragraph : Arial : 3(12pt)·-·-·T·?· 5 3 4 5 6 8

Explanation / Answer

Solution:-

27)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P < 0.55
Alternative hypothesis: P > 0.55

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.05946
z = (p - P) / S.D

z = 0.841

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 0.841.

Thus, the P-value = 0.20

Interpret results. Since the P-value (0.20) is more than the significance level (0.05), we have to accept the null hypothesis.

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