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Solution :
The null and alternative hypothesis is given by
Assume that the level of significance, α = 0.05
The t-test statistics is, t = 1.484
Degree of freedom :
df = n - 1
= 14 - 1
= 13
The P-value is given by
P-value = 2P (T>|t|)
= 2P(T>|1.484|)
= 2P(T>1.484)
= 2(=T.DIST.RT(1.484,13))
= 0.05