MGMT2263
Worksheet #1
1) A company believes its site receives significantly fewer than 300 hits per day on average. For a sample of 60 days, the average was 283 hits with a standard deviation of 58.
a) Test the hypothesis at a 5% level of significance? (Z = -2.27; conclude the site receives fewer than 300 hits per day on average)
b) If the level of significance were 1% instead, would the conclusion be different? (p-value = 0.0116; since p-value > 1%, conclusion would be different)
c) What would be the power of the test at a 5% level of significance if the average number of hits per day is 283 based on the data provided? (73.57%)
d) What would the alternative mean need to be at a 5% level of significance if they wanted the power to be 99%? Round the value to 1 decimal. (270.3)
2) Based on historical data, the life span of an electronic chip used in a PC is normally distributed with a standard deviation of 20 hours. A sample of 40 chips has an average life span of 625.4 hours.
a) Construct a 90% confidence interval for the mean life of a chip and interpret what it means. (620.2 < m < 630.6; We are 90% confident that the average chip life is between 620.2 and 630.6 hours)
b) If we test the hypothesis that the average life of a chip is 633 hours, what conclusion would we reach using a 10% level of significance? (conclude the average life is not 633 hours since this value does not fall in the 90% confidence interval)
c) What is the highest level of significance in which we would not reject the null hypothesis given the evidence? (1.64%)
d) What would be the power of the test at a 10% level of significance if the average life span of a chip is 625 hours? (81.06%)
3) A weight loss clinic wants to see if the average college student weighs more than 175 pounds. A sample of 70 students had an average weight of 182 pounds with a standard deviation of 28 pounds.
a) Does the average student weigh significantly more than 175 pounds? Test at a 3% level of significance. (Z = 2.09; conclude the average college student weighs more than 175 pounds)
b) Construct a 95% confidence interval of the weight of the average college student. Interpret the interval. (175.44 < m < 188.56; With 95% confidence, the average college student weighs between 175 and 189 pounds)
c) What would be the power of the test at a 3% level of significance if the average weight of students is 182 pounds? (58.32%)
d) What would the power of the test increase to if the level of significance were increased to 10%? (79.1%)
4) In an effort to control costs, a quality control inspector is interested in whether the mean number of ounces of sauce dispensed by bottle-filling machines differs from 16 ounces. From the bottling process, the inspector collects the following measurements: 16.3, 16.2, 15.8, 15.4, 16.0, 15.6, 15.5, 16.1, 15.9, 16.1.
a) Test at a 5% level of significance that the bottle-filling machines need adjusting. Assume that the distribution of the amount dispensed is normally distributed. (t = -1.1326; conclude the machine does not need adjusting)
b) Is there a level of significance between 1% and 10% in which the opposite conclusion would have been reached? (p-value is greater than 0.1; since this provide strong evidence for Ho, the answer is no)
5) According to a money management firm, the standard deviation of the daily closing prices on the NYSE has been 13%. The chair of a mutual fund company wishes to determine if the standard deviation of equity mutual funds differs from 13%. The assumption is that returns on mutual funds are normally distributed. A sample of 25 equity funds shows a standard deviation of 9.8%.
a) Test the hypothesis at a 5% level of significance.(c2 = 13.6388; conclude the standard deviation is not significantly different from 13%)
b) Construct a 95% confidence interval of the standard deviation of equity mutual funds. Show why you would have reached the same conclusion as in part a.
(7.6521% < s < 13.6333%; since s = 13% falls in this interval we would not reject Ho at a 5% level of significance)
c) Is there a level of significance between 1% and 10% in which the opposite verdict would have been reached? (If the level of significance were 10%, the critical value on the left side would be 13.8484 and the test statistic would fall in the rejection region)
6) A manager at National Insurance believes that more than half of claims are due to speeding. From a random sample of 75 claims, 40 were found to be associated with speeding.
a) Test the manager’s belief at the 5% level of significance. (Z = 0.58; conclude that no more than half of claims are due to speeding)
b) Suppose a level of significance had not been chosen. Why would the same conclusion be reached? (p-value = 0.281; since 28.1% > 10%, we would not reject Ho and reach the same conclusion)
c) Construct a 95% confidence interval of the percentage of claims that are associated with speeding. Interpret the interval. (42.04% < p < 64.62%; with 95% confidence, the average percentage of claims associated with speeding ranges from 42.04% to 64.62%)
d) What would be the probability of a Type II error at a 5% level of significance if the actual percentage of claims due to speeding is 53%? (87.08%)
7) A researcher wants to determine if the percentage of executives using a particular brand of PDA has significantly changed from 25%. A survey of 800 executives shows that 236 of them use this brand of PDA.
a) Test the hypothesis at a 4% level of significance. (Z = 2.94; conclude the percentage has significantly changed from 25%)
b) Suppose a level of significance had not been chosen. Why would the same conclusion be reached? (p-value = 0.0016; p-value < 1%)
c) What would be the power of the test if the percentage of executives using this brand of PDA is 30% at a 4% level of significance? (87.29%)