In: Statistics and Probability
Skagen-Tvosky Clock Manufacturers, LLC claim that their wall clocks on average will neither gain nor lose time throughout the week. A sample of 18 wall clocks showed the following gains (+) or losses (-) in seconds per week. Is it reasonable to conclude that the mean gain or loss in time for the watches is 0?
0.10 |
0.10 |
0.40 |
-0.32 |
-0.30 |
-0.23 |
-0.20 |
0.25 |
-0.10 |
-0.37 |
-0.61 |
0.20 |
-0.30 |
-0.64 |
0.40 |
-0.20 |
-0.68 |
-0.40 |
A.) |
State the decision rule for 0.05 significance level. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.) |
Reject H0: μ = 0 if t < ____________ | or t > ________ |
B.) |
Compute the Value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) |
Value of the test statistic: |
_____________ |
C.) What is the P-Value?
The sample size is n=18. The provided sample data along with the data required to compute the sample mean Xbar and sample variance s2 are shown in the table below:
Gain | Gain2 | |
0.10 | 0.01 | |
0.10 | 0.01 | |
0.40 | 0.16 | |
-0.32 | 0.1024 | |
-0.30 | 0.09 | |
-0.23 | 0.0529 | |
-0.20 | 0.04 | |
0.25 | 0.0625 | |
-0.10 | 0.01 | |
-0.37 | 0.1369 | |
-0.61 | 0.3721 | |
0.20 | 0.04 | |
-0.30 | 0.09 | |
-0.64 | 0.4096 | |
0.40 | 0.16 | |
-0.20 | 0.04 | |
-0.68 | 0.4624 | |
-0.4 | 0.16 | |
Sum = | -2.9 | 2.409 |
The sample mean Xbar is computed as follows:
Now applying T-test for the analysis
The provided sample mean is \bar X = -1.61Xˉ=−1.61 and the sample standard deviation is s = 0.338s=0.338, and the sample size is n = 18n=18.
Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: μ = 00
Ha: μ ≠ 0
This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.
Rejection Region
Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is tc=2.11.
The rejection region for this two-tailed test is R={t:∣t∣>2.11}
Here critical value is calculated using t distribution with 17 degrees of freedom with two-tailed distribution.
So reject H0 if
t < -2.11 or t>0.211
Test Statistics
The decision about the null hypothesis
Since it is observed that |t| = 20.209 ∣t∣=20.209>tc=2.11, it is then concluded that the null hypothesis is rejected.
Using the P-value approach
The p-value is p=0.0, and since p=0.00 <0.05, it is concluded that the null hypothesis is rejected.
Here p-value is calculated using t distribution using 17 df
p-value = P[∣t∣>20.209]
p-value = 2*P[t>20.209]
p-value = 2*0.00..
p-value = 0.00..
Conclusion
It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is different than 0, at the 0.05 significance level. Hence, it is reasonable to conclude that the mean gain or loss in time for the watches is 0?