Fitchburg State University Statistics Worksheet

Problem 1. (6 points)Dynacel batteries last 13.29 hours in a standard flashlight, on average. Dynacel engineers have developed a new manufacturing process that they hope will increase the mean life of their batter Remaining time: 29:49 (min:sec)
sample of batteries to determine if the new process has indeed increased the average life of the batteries.
For the test, their hypotheses are:
Use mu for u, xbar for X, and sigma for o, as needed.
Use , and != for #, as needed.
Enter percents as decimals, without a leading 0. For example, enter 50% as .50.
Ho:
Ha:
The test is
30 batteries made by the new process are randomly selected and tested.
To test at 1% significance, what is the critical t-score (3 decimal places)?
The new batteries had a mean life of 14.98 hours with a standard deviation of 3.7 hours.
Calculate the test statistic for the sample (round to 3 decimal places).
What is the test conclusion?
A. Accept the alternative hypothesis: The test shows that the mean is 14.98 hours.
B. Reject the alternative hypothesis: The test provides sufficient evidence to show the mean is not greater than 13.29 hours.
C. Reject the null hypothesis: The test provides evidence that the mean is greater than 13.29 hours.
D. Accept the null hypothesis: The test shows that the mean is 13.29 hours.
E. Don’t reject the null hypothesis: The test does not provide sufficient evidence to prove the mean is greater than 13.29 hours.
Note: You can earn partial credit on this problem.
Problem 2. (5 points)
Remaining time: 29:32 (min:sec)
FastGro is working on a “new and improved” fertilizer, Formula 1. To test whether the new formula actually produces more plant growth than the old fertilizer, we randomly divide seeds into two groups. One group gets the old
formula, while the other group gets the new formula. After a specified time, the plants are weighed. The old-formula weights are in the Formula_0 variable, while the new-formula weights are in the Formula_1 variable. The
hypotheses and RStudio t-test results are shown below:
Ho: M1 = MO
Ha:Mi > Mo
> t.test(Formulal, Formula), mu=0, type=”greater”)
Welch Two Sample t-test
data: Formula_1 and Formula_0
t = 1.9039, df = 13.8, p-value = 0.07799
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.06947581 1.15447581
sample estimates:
mean of x mean of y
5.2035
4.661
t=
p-value
At 10% significance, what is the test conclusion?
A. The Formula_1 mean, 5.2035, is greater than the Formula_0 mean, 4.661, so reject the null hypothesis. Formula 1 increases plant growth.
B. The p-value is large, so accept the null hypothesis. The test shows that the means are the same, so Formula 1 does not increase plant growth.
C. The p-value s 10%, so reject the null hypothesis. The test provides evidence that Formula 1 increases plant growth.
D. The p-value is less than the t-score, so we reject the alternative hypothesis. Formula 1 does not increase plant growth.
E. The p-value > 10%, so we cannot reject the null hypothesis. The test does not provide sufficient evidence to prove that Formula 1 increases plant growth.
At 10% significance, what is the test conclusion?
A. The Formula_1 mean, 5.2035, is greater than the Formula_0 mean, 4.661, so reject the null hypothesis. Formula 1 increases plant growth.
B. The p-value is large, so accept the null hypothesis. The test shows that the means are the same, so Formula 1 does not increase plant growth.
C. The p-value s 10%, so reject the null hypothesis. The test provides evidence that Formula 1 increases plant growth.
D. The p-value is less than the t-score, so we reject the alternative hypothesis. Formula 1 does not increase plant growth.
E. The p-value > 10%, so we cannot reject the null hypothesis. The test does not provide sufficient evidence to prove that Formula 1 increases plant growth.
At 5% significance, what is the test conclusion?
A. The Formula_1 mean, 5.2035, is greater than the Formula_0 mean, 4.661, so reject the null hypothesis. Formula 1 increases plant growth.
B. The p-value > 5%, so we cannot reject the null hypothesis. The test does not provide sufficient evidence to prove that Formula 1 increases plant growth.
C. The p-value s 5%, so reject the null hypothesis. The test provides evidence that Formula 1 increases plant growth.
D. The p-value is large, so accept the null hypothesis. The test shows that the means are the same, so Formula 1 does not increase plant growth.
O E. The p-value is less than the t-score, so we reject the alternative hypothesis. Formula 1 does not increase plant growth.
At 1% significance, what is the test conclusion?
A. The Formula_1 mean, 5.2035, is greater than the Formula_0 mean, 4.661, so reject the null hypothesis. Formula 1 increases plant growth.
B. The p-value > 1%, so we cannot reject the null hypothesis. The test does not provide sufficient evidence to prove that Formula 1 increases plant growth.
C. The p-value s 1%, so reject the null hypothesis. The test provides evidence that Formula 1 increases plant growth.
D. The p-value is less than the t-score, so we reject the alternative hypothesis. Formula 1 does not increase plant growth.
E. The p-value is large, so accept the null hypothesis. The test shows that the means are the same, so Formula 1 does not increase plant growth.
Note: You can earn partial credit on this problem.
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Dynacel batteries last 13.29 hours in a standard flashlight, on average. Dynacel engineers have developed a new manufacturing process that they hope will increase the mean life of their batteries. They will test a random
sample of batteries to determine if the new process has indeed increased the average life of the batteries.
Remaining time: 26:07 (min:sec)
For the test, their hypotheses are:
3
Use mu for u, xbar for X, and sigma for o, as needed.
Use , and != for #, as needed.
Enter percents as decimals, without a leading 0. For example, enter 50% as .50.
HO:
Ha:
The test is
30 batteries made by the new process are randomly selected and tested.
To test at 1% significance, what is the critical t-score (3 decimal places)?
The new batteries had a mean life of 14.98 hours with a standard deviation of 3.7 hours.
Calculate the test statistic for the sample (round to 3 decimal places).
What is the test conclusion?
A. Accept the alternative hypothesis: The test shows that the mean is 14.98 hours.
B. Reject the alternative hypothesis: The test provides sufficient evidence to show the mean is not greater than 13.29 hours.
C. Reject the null hypothesis: The test provides evidence that the mean is greater than 13.29 hours.
D. Accept the null hypothesis: The test shows that the mean is 13.29 hours.
O E. Don’t reject the null hypothesis: The test does not provide sufficient evidence to prove the mean is greater than 13.29 hours.