Question

A student suspected the average cost of a Saturday night date was no longer $30.00.TO test her hypothesis she randomly selected 16 men from dormitory and asked how much they spent on a date last Saturday. She found that the average cost was $31.17. The standard deviation of sample was $5.51. At α=0.05, is there enough evidence to support her claim?

Answer 1

Null hypothesis H₀: μ = 30

Alternative hypotheses H₁: μ > 30

n = 16

Mean X = 31.17

Standard deviation σ = 5.51

Rejection Region

This is right tailed test, for α = 0.05 and d.f. = 15

Critical value of t is 1.753

Hence reject H₀ if t > 1.753

$t = \frac{X - μ}{\frac{σ}{\sqrt{n}}} \\ t = \frac{31.17 – 30}{\frac{5.51}{\sqrt{16}}} \\ t = 0.849 \\ t < t_{crit}$

fail to reject null hypothesis

There is enough evidence to support her claim.