The test statistic is -0.46 and the p-value is gotten as 0.3340
<h3>How to Calculate the P-value?</h3>
The question is not complete because the data from the sample is 105, 107, 117, 106, 110.
Mean of sample is;
Mean = (105 + 107 + 117 + 106 + 110)/5
Mean; x' = 109
From online standard deviation calculator, the standard deviation of the sample is; S = 4.848
Now, the claim is that your team scored an average significantly less than 110 points. Thus, let's define the hypothesis as;
Null Hypothesis; μ = 110
Alternative Hypothesis; μ < 110
standard error of the mean; S_m = S/√n
S_m = 4.848/5
S_m = 2.17
Thus, the t-statistic is;
t = (x' - μ)/S_m
t = (109 - 110)/2.17
t = -0.46
Degree of freedom for this sample size is;
D.F = N - 1 =5 - 1 = 4
From online p-value from t-score calculator, we have;
p-value = 0.3340
The p-value is greater than the significance value and as such we fail to reject the null hypothesis and conclude that there is not enough evidence to support the claim that your team scored an average significantly less than 110 points.
Read more about p-value at; brainly.com/question/4621112
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