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36 16. When mice are subjected to stress their endogenous opioid system is activated and the mice increase food consumption i
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Answer #1

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:

n1 0.9500 0.6875 1.412 10.00-1)0.5142422+(8.00-1)0.10606621 10.00+8.00-2 10.00 8.00

(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

normaldistributiongrapher.php?mean=0&sig

Please let me know in comments if anything is not clear. Will reply ASAP. Please do upvote if satisfied!!

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