Question

The director of the admissions at a large university advises parents of incoming students about the cost of textbooks during a typical semester. He selected a sample of 100 students and recorded their textbook expenses for the semester. He then computed a sample mean cost of $315.40 and a sample standard deviation of $43.20

Required:
Using the 0.10 level of significance, is there evidence that the population mean is above $300?

Answer 1

Answer:

The calculated value t = 3.5648> 1.9842 at 0.1 level of significance

The null hypothesis is rejected

The mean of the sample  does not come from the Population

Step-by-step explanation:

Step(i):-

Given that sample size 'n' = 100

Mean of the sample x⁻ = 315.40

The standard deviation of the sample (s) = 43.20

Level of significance = 0.1

Mean of the Population = 300

Null hypothesis: H₀: x₀ = μ

Alternative Hypothesis: H₁: x₀ ≠ μ

Step(ii):-

Test statistic

         $t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }$

         $t = \frac{315.40-300 }{\frac{43.20}{\sqrt{100} } }$

       t = 3.5648

Degrees of freedom = n-1 = 100-1 =99

$t_{\frac{0.1}{2} , 99} = t_{0.05,99} = 1.9842$

The calculated value t = 3.5648> 1.9842 at 0.1 level of significance

The null hypothesis is rejected

The mean of the sample  does not come from the Population