SOLUTION: The commercial for the new Mad meat Man Barbeque claims that it takes 20 minutes for assembly.
A consumer advocate thinks that the assembly time is higher than 20 minutes.
The ad
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-> SOLUTION: The commercial for the new Mad meat Man Barbeque claims that it takes 20 minutes for assembly.
A consumer advocate thinks that the assembly time is higher than 20 minutes.
The ad
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Question 1183058: The commercial for the new Mad meat Man Barbeque claims that it takes 20 minutes for assembly.
A consumer advocate thinks that the assembly time is higher than 20 minutes.
The advocate surveyed 15 randomly people who purchased the Meat Man Barbeque and found that their average time was 21.3 minutes.
The standard deviation for this survey group was 2.6 minutes.
What can be concluded at the a = 0.10 level of significance?
a) For this study, we should use = t-test for a population mean
b) The null and alternative hypotheses would be:
Ho: u = ________
c) The test statistic t = _______
To 3 decimals
d) The p-value = ______
To 4 decimals Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! test mean is 20 minutes.
sample mean is 21.3 minutes with standard deviation of 2.6 minutes.
sample size is 15.
test is whether sample mean is higher than test mean.
this indicates a one tailed confidence interval with the tail on the high end.
null hypothesis is assembly time is 20 minutes.
alternate hypothesis is assembly time is greater than 20 minutes.
standard error = standard deviation / sqrt(sample size) = 2.6/sqrt(15) = .671317 rounded to 6 decimal places.
t-score = (x - m) / s
x is the sample mean
m is the test mean
s is the standard error.
t = (21.3 - 20) / .671317 = 1.936 rounded to 3 decimal places.
degrees of freedom = 15 - 1 = 14.
probability of getting a t-score greater than 1.936492 with 14 degrees of freedom is .0367.
critical p-value at 14 degrees of freedom is .10.
test p-value is lower than this.
test is significant.
the alternate hypothesis that states that the assembly time is greater than 20 minutes is accepted.
test statistic is t(14) = 1.936 rounded to 3 decimal places.
p-value = .0367
your inputs for this test are:
t = 1.936
degrees of freedom = 14
test is one tail
critical p-value is .10
note that if the test is > or <, then the the test is one tail and the critical p-value is .10 on only one end, and if the test is two tail, the critical p-value is cut in half = .05 and placed on both ends of the confidence interval.
