An investigator wants to assess whether the mean (mu) = the average weight of passengers flying on small planes exceeds the FAA guideline of average total weight of 185 pounds (passenger weight including shoes, clothes, and carry-on). Suppose that a random sample of 25 passengers showed an average total weight of 200 pounds with a sample standard deviation of 59.5 pounds. Assume that passenger total weights are normally distributed. Which of the following is an appropriate conclusion (alpha= .05)? Group of answer choicesnull hypothesis should be revisedthe average total weight of all passengers appears to be not greater than 185 pounds.the average total weight of all passengers appears to be greater than 185 pounds.
Question
An investigator wants to assess whether the mean (mu) = the average weight of passengers flying on small planes exceeds the FAA guideline of average total weight of 185 pounds (passenger weight including shoes, clothes, and carry-on). Suppose that a random sample of 25 passengers showed an average total weight of 200 pounds with a sample standard deviation of 59.5 pounds. Assume that passenger total weights are normally distributed. Which of the following is an appropriate conclusion (alpha= .05)? Group of answer choicesnull hypothesis should be revisedthe average total weight of all passengers appears to be not greater than 185 pounds.the average total weight of all passengers appears to be greater than 185 pounds.
Solution
To answer this question, we need to conduct a one-sample t-test to compare the sample mean (200 pounds) to the population mean (185 pounds) under the null hypothesis that the population mean is 185 pounds.
Step 1: State the hypotheses. The null hypothesis is that the average total weight of all passengers is 185 pounds. The alternative hypothesis is that the average total weight of all passengers is greater than 185 pounds.
Step 2: Calculate the test statistic. The formula for the test statistic in a one-sample t-test is (sample mean - population mean) / (sample standard deviation / sqrt(sample size)). Plugging in the given values, we get (200 - 185) / (59.5 / sqrt(25)) = 1.27.
Step 3: Determine the critical value for a one-tailed t-test with 24 degrees of freedom (sample size - 1) at the 0.05 significance level. The critical value is approximately 1.711.
Step 4: Compare the test statistic to the critical value. Since 1.27 is less than 1.711, we do not reject the null hypothesis.
Therefore, the appropriate conclusion is that the average total weight of all passengers appears to not be greater than 185 pounds.
Similar Questions
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