The data is:
| Loud noise | Normal noise | |
| 1.05 | 0.60 | |
| 0.80 | 0.80 | |
| 1.15 | 0.75 | |
| 0.25 | 0.65 | |
| 1.60 | 0.70 | |
| 0.30 | 0.60 | |
| 0.95 | 0.55 | |
| 1.85 | 0.85 | |
| 0.55 | ||
| 1.00 | ||
| Count | 10.00 | 8.00 |
| Mean | 0.9500 | 0.6875 |
| StDev | 0.514242 | 0.106066 |
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: μ1 = μ2
Ha: μ1 > μ2
This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.
(2) Rejection Region
Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=16. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:
Hence, it is found that the critical value for this right-tailed test is tc=1.746, for α=0.05 and df=16.
The rejection region for this right-tailed test is R={t:t>1.746}.
(3) Test Statistics
Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis
Since it is observed that t=1.412≤tc=1.746, it is then concluded that the null hypothesis is not rejected.
Using the P-value approach: The p-value is p=0.0886, and since p=0.0886≥0.05, it is concluded that the null hypothesis is not rejected.
(5) Conclusion
It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1 is greater than μ2, at the 0.05 significance level.
Graphically
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