A sample of colored candies was obtained to determine the weights of different colors. The ANOVA table is shown below. It is known that the population distributions are approximately normal and the variances do not differ greatly. Use a 0.025 significance level to test the claim that the mean weight of different colored candies is the same. If the candy maker wants the different color populations to have the same mean weight, do these results suggest that the company has a problem requiring corrective action? Source: DF: Ss: MS: Test Stat, F: Critical F: P-Value: 0.001 0.001 Treatment: 5 0.005 0.6049 2.7361 0.6963 Error: 77 0.077 Total: 82 0.082 Should the null hypothesis that all the colors have the same mean weight be rejected? O A. No, because the P-value is greater than the significance level. O B. Yes, because the P-value is greater than the significance level. OC. No, because the P-value is less than the significance level. O D. Yes, because the P-value is less than the significance level. Does the company have a problem requiring corrective action? O A. Yes, because it is not likely that the candies do not have equal mean weights. O B. No, because it is likely that the candies do not have equal mean weights. OC. Yes, because it is likely that the candies do not have equal mean weights. O D. No, because it is not likely that the candies do not have equal mean weights.
A sample of colored candies was obtained to determine the weights of different colors. The ANOVA table is shown below. It is known that the population distributions are approximately normal and the variances do not differ greatly. Use a 0.025 significance level to test the claim that the mean weight of different colored candies is the same. If the candy maker wants the different color populations to have the same mean weight, do these results suggest that the company has a problem requiring corrective action? Source: DF: Ss: MS: Test Stat, F: Critical F: P-Value: 0.001 0.001 Treatment: 5 0.005 0.6049 2.7361 0.6963 Error: 77 0.077 Total: 82 0.082 Should the null hypothesis that all the colors have the same mean weight be rejected? O A. No, because the P-value is greater than the significance level. O B. Yes, because the P-value is greater than the significance level. OC. No, because the P-value is less than the significance level. O D. Yes, because the P-value is less than the significance level. Does the company have a problem requiring corrective action? O A. Yes, because it is not likely that the candies do not have equal mean weights. O B. No, because it is likely that the candies do not have equal mean weights. OC. Yes, because it is likely that the candies do not have equal mean weights. O D. No, because it is not likely that the candies do not have equal mean weights.
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.4: Distributions Of Data
Problem 19PFA
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Transcribed Image Text:A sample of colored candies was obtained to determine the weights of different colors. The ANOVA table is shown below. It is known that the population distributions
are approximately normal and the variances do not differ greatly. Use a 0.025 significance level to test the claim that the mean weight of different colored candies is the
same. If the candy maker wants the different color populations to have the same mean weight, do these results suggest that the company has a problem requiring
IS
corrective action?
Source:
DF:
s:
MS:
Test Stat, F: Critical F: P-Value:
0.6049
0.001
0.001
Treatment: 5
0.005
2.7361
0.6963
Error:
77
0.077
Total:
82
0.082
Should the null hypothesis that all the colors have the same mean weight be rejected?
O A. No, because the P-value is greater than the significance level.
or
B. Yes, because the P-value is greater than the significance level.
O C. No, because the P-value is less than the significance level.
O D. Yes, because the P-value is less than the significance level.
Does the company have a problem requiring corrective action?
it
O A. Yes, because it is not likely that the candies do not have equal mean weights.
SH
is
O B. No, because it is likely that the candies do not have equal mean weights.
O C. Yes, because it is likely that the candies do not have equal mean weights.
O D. No, because it is not likely that the candies do not have equal mean weights.
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