Step 4 make the decision and state the conclusion for

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Step 4: Make the decision and State the conclusion for the test If the test statistic falls within the rejection region, we reject H 0 . We have sufficient/significant evidence to support the alternative hypothesis. If the test statistic does NOT fall within the rejection region, then we fail to reject H 0 . There is insufficient/no significant evidence to support the alternative hypothesis. STAT3502 CU 24 / 41
t test for one population mean Inference for one population mean when σ is Unknown Example – Fast Food A fast food restaurant claims that the average waiting time in their drive through is less than one minute. We record the drive-thru waiting times for a random sample of 31 customers. The sample average time is 54.9 seconds and the sample standard deviation is 17.6 seconds. Waiting times are known to follow a normal distribution. Conduct a hypothesis test to examine the significance of the restaurant’s claim at the 1% level of significance. Solutions: From the question, we have n = 31, x = 54 . 9, s = 17 . 6, and α = 0 . 01. The value of σ is given. Therefore, we use a t procedure. STAT3502 CU 25 / 41 t test for one population mean Inference for one population mean when σ is Unknown Example – Fast Food Let μ be the average waiting time in the drive though. step 1. Stating Hypotheses H 0 : μ = 60 vs. H 1 : μ < 60 step 2. Finding the comparative number Given α = 0 . 01 and n = 31, we look at Table A-3 with d.f. = n - 1 = 30 We found that - t α = - 2 . 457. STAT3502 CU 26 / 41 t test for one population mean Inference for one population when σ is Unknown Example – Fast Food Rejection Rule: We reject H 0 at level of significance 1% if t test statistic -2.457. We fail to reject H 0 at level of significance 1% if t test statistic > -2.457. Step 3: Calculate test statistic t = x - μ 0 s / n = 54 . 9 - 60 17 . 6 / 31 = - 1 . 61 Step 4: Make the decision and State the conclusion for the test Since - 1 . 61 > - 2 . 457, our t test statistic does NOT fall within the rejection region. Fail to reject H 0 at the level of significance 1%. Therefore, we have insufficient evidence to conclude that the true mean waiting time in the drive-thru is less than one minute. STAT3502 CU 27 / 41 t test for one population mean Inference for one population – Grooms Example z v.s. t In 1968, census results indicated that the age at which American men first married had a mean of 23.3 years and a standard deviation of 4.8 years. We take a random sample of 28 men who married for the first time in 2008. Their average age was 25.2 years and the standard deviation of their ages was 4.6 years. We assume that ages of first-time grooms are normally distributed. Conduct a hypothesis test to determine whether the mean age of first-time grooms has changed since 1968 at significant level α = 0 . 05. STAT3502 CU 28 / 41 t test for one population mean Inference for one population – Grooms Example z v.s. t Solution: Let μ be the mean age of first-time grooms in 2008. Let X be the age of American men married for the first time. From the question, we have X N ( μ = 23 . 3 , σ = 4 . 8 ) , n = 28, x = 25 . 2 and s = 4 . 6. Because σ is known , we choose a z procedure for this hypothesis testing using p-value method. STAT3502 CU 29 / 41 t test for one population mean

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