LA\'s Department of Transportation and Development (DOTD is repairing a stretch
ID: 3232153 • Letter: L
Question
LA's Department of Transportation and Development (DOTD is repairing a stretch of I-10 near LSU. DOTD wants to ensure the surface will be effective for the volume of heavy traffic in the area. State weight stations report that heavy trailers average 72 per hour (based on I-12 data). DOTD believes that the average for the area to be repaired is larger since it must also account for I-10 traffic. A random sample scheme was employed to collect data for 1 month in 50 1-hr periods, results were: x^- = 74.1, s = 13.3 Does the sample support DOTD's claims? How would the answer change if n = 100? How would the answer change if alpha = 0.05? What are the chances of making a Type I and Type II Error?Explanation / Answer
Solution:-
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 72
Alternative hypothesis: > 72
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 1.88
DF = n - 1 = 50 - 1
D.F = 49
t = (x - ) / SE
t = 1.12
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a t statistic test statistic of 1.12. We use the t Distribution Calculator to find P(t > 1.12) = 0.134
Interpret results. Since the P-value (0.134) is greater than the significance level (0.10), we cannot reject the null hypothesis.
2) If n = 100
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 72
Alternative hypothesis: > 72
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 1.33
DF = n - 1 = 100 - 1
D.F = 99
t = (x - ) / SE
t = 1.58
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a t statistic test statistic of 1.58. We use the t Distribution Calculator to find P(t > 1.58) = 0.059
Interpret results. Since the P-value (0.059) is less than the significance level (0.10), we have to reject the null hypothesis.
c) If alpa = 0.05
The observed sample mean produced a t statistic test statistic of 1.58. We use the t Distribution Calculator to find P(t > 1.58) = 0.059
Interpret results. Since the P-value (0.059) is greater than the significance level (0.05), we have to accept the null hypothesis.
d) The chances of making Type I error is 0.05.
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