Download Hypothesis Testing for Means Answers

Hypothesis Testing for Means Answers 1. An insurance company is reviewing its current policy rates. When originally setting the rates they believed that the average claim amount was $1,800. They are concerned that the true mean is actually higher than this, because they could potentially lose a lot of money. They randomly select 40 claims, and calculate a sample mean of $1,950. Assuming that the standard deviation of claims is $500, and set α = 0.05, test to see if the insurance company should be concerned. Let µ = Average Claim Amount Ho: µ = $1800 Ha: µ > $1800 I will perform a one-­‐sample test for averages. $1800 $1950 X = $1950 z =
1950 −1800
= 1.897 500
40
z* = 1.645 p = 0.0289 α = 0.05 Reject Ho because p <0.05 and z > 1.645. Data is statistically significant. We have strong evidence to conclude that average claim amount is more than $1800 and that the insurance company should increase their policy rates 2. A sample of 40 sales receipts from a grocery store has a mean of $137 and a standard deviation of $30. Use these values to test whether or not the mean is sales at the grocery store are different from $150. Let µ = Average Sale per customer Ho: µ = $150 Ha: µ ≠ $150 I will perform a one-­‐sample test for averages. $137 $150 $163 150 −137
= −2.741 30
40
X = $137 z=
z* = 1.96 p = 0.00613 α = 0.05 Reject Ho because p <0.05 and z > 1.96 . Data is statistically significant. We have strong evidence to conclude that average sale receipt is not equal to $150 3. You are the manager of the packaging process at a cereal manufacturing plant. You want to determine if the cereal filling process is in control. The process requires no corrective action if the correct amount of cereal per box is 368 grams. To study this, you decide to take a random sample of 25 boxes, weigh each one, and then evaluate the difference between the sample statistic and the hypothesized population parameter by comparing the mean weight from the sample to the expected mean of 368 grams specified by the company. The sample mean is 372.5 and the process standard deviation is 15. Is there evidence that the weight is different from 368 grams? Let µ = Average amount of cereal in box Ho: µ = 368 grams Ha: µ ≠ 368 grams I will perform a one-­‐sample test for averages. 363.5 368 372.5 372.5 − 368
= 1.5 15
25
X = 372.5grams z =
z* = 1.96 p = 0.134 α = 0.05 Do not Reject Ho because p > 0.05 and z <1.96. Data is not statistically significant. We do not have sufficient evidence to conclude that average amount of cereal is different from 368 grams. Therefore we cannot conclude that the filling process is not in control 4. A manufacturer of steel rods considers that the manufacturing process is working properly if the mean length of the rods is 8.6. The standard deviation of these rods always runs about 0.3 inches. Suppose a random sample of size n = 36 yields an average length of 8.7 inches. Should the manufacturer conclude the process is working properly or improperly? Use the .05 level of significance Let µ = Average length of steel rods Ho: µ = 8.6 inches Ha: µ ≠ 8.6 inches I will perform a one-­‐sample test for averages. 8.5 8.6 8.7 8.7 − 8.6
= 2 0.3
36
X = 8.7inches z=
z* = 1.96 p = 0.0455 α = 0.05 Reject Ho because p < 0.05 and z >1.96 . Data is statistically significant. We have strong evidence to conclude that the rod size is not 8.6 inches and that the manufacturing process is not working correctly. 5. New tires manufactured by a company outside Ocala, Fl. , are designed to provide a mean life expectancy of at least 40,000 miles. The tire rating has a standard deviation of 1000 miles. A test with 30 tires shows a sample mean of 39,600 miles. Using a 0.02 level of significance, test whether or not there is sufficient evidence to reject the claim of a mean of at least 40,000 miles. Let µ = Average life expectancy of tire Ho: µ = 40000 miles Ha: µ < 40000 miles I will perform a one-­‐sample test for averages. 39600 40000 39600 − 40000
= −2.19 p = 0.0142 1000
30
X = 39600 miles z =
z* = 2.054 α = 0.02 Reject Ho because p <0.02 and |z| >2.054. Data is statistically significant. We have strong evidence to conclude that average life expectancy of tires is less than 40000 miles 6. The Public Health Service publishes the "Annual Data Tabulations, Continuous Air Monitoring Projects", which some years ago indicated that a large Midwestern city had an annual mean level of sulfur dioxide of 0.12 (concentration in parts per million). To change this concentration, many steel mills and other manufacturers installed antipollution equipment. In order to study the effects of these efforts to reduce pollution 36 random checks were made throughout the year. It was found that the sample mean pollution sulfur dioxide level was 0.115 with a sample standard deviation of 0.03. At the 0.05 level of significance, does this evidence suggest that there has been a change in the sulfur dioxide level in this city? Let µ = Average Sulfur Dioxide Level(ppm) Ho: µ = 0.12 ppm Ha: µ < 0.12 ppm I will perform a one-­‐sample test for averages. 0.115 0.12 X = 0.115 ppm 0.05 z =
0.115 − 0.12
= −1 0.03
36
z* = 1.645 p = 0.159 α = Do not Reject Ho because p >0.05 and |z| <1.645. Data is not statistically significant. We do not have strong evidence to conclude that average sulfur dioxide level is less than 0.12 ppm. Therefore we do not have evidence that the anti-­‐pollution equipment is successful at reducing sulfur dioxide levels. 7. A machine is set to fire 30.00 decigrams of chocolate pellets into a box of cake mix as it moves along the production line. Of course, there is some variation in the weight of the pellets. A sample of 36 boxes of mix revealed that the average weight of the chocolate pellets was 30.18 decigrams with a sample standard deviation of 0.50 decigrams. At the 0.05 level of significance, is the sampled weight of the chocolate pellets of 30.18 decigrams suggesting that the mean weight of the chocolate pellets in all cake mix boxes is higher than the set 30.00 decigrams? Let µ = Average number of chocolate pellets Ho: µ = 30 decigrams Ha: µ > 30 decigrams I will perform a one-­‐sample test for averages. 30 30.18 30.18 − 30
= 2.16 0.5
36
X = 30.18decigrams z=
z* = 1.645 p = 0.0154 α = 0.05 Reject Ho because p <0.05 and z >1.645 . Data is statistically significant. We have strong evidence to conclude that machine is dispersing more than 30 decigrams of chocolate and that the machine is not working properly 8. A survey found that the average hotel room rate in New Orleans is $88.42 and the average room rate in Phoenix is $80.61. Assume that the data were obtained from two samples of 50 hotels each and that the standard deviations were $5.62 and $4.83, respectively. At α= 0.05, can it be concluded that there is a significant difference in the rates? Let µN = Average Room Rate in New Orleans µP = Average Room Rate in Phoenix Ho: µN = µP (µN -­‐ µP = 0) Ha: µN ≠ µP (µN -­‐ µP ≠ 0) I will perform a two-­‐sample test for averages -­‐7.81 0 7.81 X N − X P = $7.81 z =
7.81
5.62 2 4.832
+
50
50
= 7.452 p ≈ 0.00 z* = 1.96 α = 0.05 Reject Ho because p < 0.05 and z >1.96. Data is statistically significant. We have strong evidence to that the average cost for a hotel room in Phoenix and New Orleans is not the same 9. A researcher hypothesizes that the average number of sports that colleges offer for males is greater than the average number of sports that colleges offer for females. Data was randomly collected from 50 colleges. The average number of male sports offered is 8.6 with a standard deviation of 3.3 and the average number of female sports offered is 7.9 with a standard deviation of 3.3 At α = 0.05, is there enough evidence to support the claim? Let µM = Average Number of Male sports per college µF = Average Number of Female sports per college Ho: µM = µF (µM -­‐ µF = 0) Ha: µM > µF (µM -­‐ µF > 0) I will perform a two-­‐sample test for averages 0 0.7 X M − X F = 0.7 z =
0.7
2
2
3.3 3.3
+
50
50
z* = 1.645 = 1.061 p = 0.144 α = 0.05 Do not Reject Ho because p > 0.05 and z < 1645. Data is not statistically significant. We do not have strong evidence to conclude that the average number of male sports offered at universities is more that the average number of female sports offered at universities 10. A researcher wishes to determine whether the salaries of professional nurses employed by private hospitals are higher than those of nurses employed by government-­‐owned hospitals. She selects a sample of nurses from each type of hospital and calculates the means and standard deviations of their salaries. At α = 0.01, can she conclude that the private hospitals pay more than the government hospitals? Let µP = Average Nurse salary at Private Hospital µG = Average Nurse salary at Government Hospital Ho: µP = µG (µP -­‐ µG = 0) Ha: µP > µG (µP -­‐ µG > 0) I will perform a two-­‐sample test for averages 0 $1400 X P − X G = $1400 z =
1400
600 2 450 2
+
10
8
= 5.654 p ≈ 0.00 z* = 2.326 α = 0.01 Reject Ho because p < 0.01 and z >2.326. Data is statistically significant. We have strong evidence to that the average salary for a nurse at a private hospital is more than the average salary for a nurse at a government 11. Universities and colleges in the United States of America are categorized by the highest degree offered. Type A institutions offer a Master's Degree, and Type B institutions offer a Baccalaureate degree. A professor, looking for a new position, wonders if the salary difference between Type A and B institutions is really significant. He finds that a random sample of 200 Type A institutions has a mean salary (for full professors) of $54,218.00, with standard deviation $8,450. A random sample of 200 Type B institutions has a mean salary (for full professors) of $46,550.00, with standard deviation $9,500 (assume that the sample standard deviations are in fact the population standard deviations). Do these data indicate a significantly lower salary at Type B institutions? Let µA = Average Salary at Type A University µB = Average Salary at Type B University Ho: µA = µB (µA -­‐ µB = 0) Ha: µA > µB (µA -­‐ µB > 0) I will perform a two-­‐sample test for averages 0 $7668 X N − X P = $7668 z =
7668
8450 2 9500 2
+
200
200
= 8.529 p ≈ 0.00 z* = 1.645 α = 0.05 Reject Ho because p < 0.05 and z >1.645. Data is statistically significant. We have strong evidence to conclude that the average salary at Type A universities is more that the average salary at Type B universities 12. Two brands of batteries are tested; X and Y, and their voltage are compared. A random sample of 27 of Brand X and 30 of Brand Y were selected. Brand X has an average voltage of 9.2 with a standard deviation of 0.3 volts and Brand Y has an average of 9.1 volts with a standard deviation of 0.1 volts. Assume both variables are normally distributed. Does the data indicate that there is a difference in the voltage of the two brands? Let µX = Average voltage for Battery X µY = Average voltage for Battery Y Ho: µX = µY (µM -­‐ µF = 0) Ha: µX ≠ µY (µX-­‐ µY ≠ 0) I will perform a two-­‐sample test for averages -­‐0.1 0 0.1 X X − X Y = 0.1volts z=
0.1
0.32 0.12
+
27
30
= 1.651 p = 0.0986 z* = 1.96 α = 0.05 Do not Reject Ho because p > 0.05 and z < 1.96. Data is not statistically significant. We do not have evidence to suggests that there is a difference in the average voltage level in the two Batteries types 13. Assume that Peet’s Coffee is looking at two possible locations for a new shop. To help make a decision they look at the number of pedestrians that go by each of the two locations in one-­‐hour segments. The results of the random sample are shown below: Location A Location B Mean number of pedestrians 421 347 Sample standard deviation 122 85 Number of hours sampled 35 50 Does the data indicate that Location A has significantly more traffic at the 1% level of significance. Let µA = Average number of pedestrians at Locations A µB = Average number of pedestrians at Locations B Ho: µA = µB (µA -­‐ µB = 0) Ha: µA > µB (µA -­‐ µB > 0) I will perform a two-­‐sample test for averages 0 74 74
z=
= 3.1 p = 0.000967 X A − X B = 74 122 2 852
+
35
50
z* = 2.326 α = 0.01 Reject Ho because p < 0.01 and z >2.326. Data is statistically significant. We have strong evidence to suggest that the average traffic at location A is more than average traffic at location B. Therefore location A is better location for the coffee shop 14. Consumer Reports tested several types of snow tires to determine how well each performed under winter conditions. When traveling on ice at 15 mph, 10 Firestone Winterfire tires had a mean stopping distance of 51 feet with a standard deviation of 8 feet. The mean stopping distance for 12 Michelin XM+S Alpine tires was 55 feet with a standard deviation of 3 feet. Can you conclude that there is a difference between the stopping distances of the two types of tires? Use α = 0.01. Assume the populations are normally distributed. Let µF = Average stopping Distance with Firestone µM = Average stopping Distance with Michelin Ho: µF = µM (µF -­‐ µM = 0) Ha: µF ≠ µM (µF -­‐ µM ≠ 0) I will perform a two-­‐sample test for averages -­‐4 0 4 X F − X M = −4 z =
−4
82 32
+
10 12
= −1.496 z* = 2.576 p = 0.135 α = 0.01 Do not Reject Ho because p > 0.01 and |z|<2.576. Data is not statistically significant. We do not have strong evidence to suggest that there is a difference in the average stopping distance between Firestone tires and Michelin tires