From the hypothesis test, it is found that that the p value is of 0.6683
At the null hypothesis, we test if the mean is still the same, that is, of 10.2 hours per week, thus:
H₀ : µ > 10.2
At the alternative hypothesis, we test if the mean has increased, that is, if it is greater than 10.2 hours per week, thus:
H₁: µ > 10.2
10 students were surveyed, so there are 9 df (degree of freedom).
The test statistic is t = 0.45, and it is a one-tailed test, as we are testing if the mean is greater than a value
Thus, p test:
X = 10.2 hours
n = 10
test statistics = 0.45
Thus, using a t-distribution calculator, the p-value of the test is of 0.6683
Pvalue = 0.6683
p value is greater is than 0.10 which indicates that the amount of time is not greater than 10.2 hours for this year.
Hence we get the value of p as 0.6683.
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