Respuesta :
Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 17
For the alternative hypothesis,
µ < 17
This is a left tailed test.
Since the population standard deviation is not given, the distribution is a student's t.
Since n = 80
Degrees of freedom, df = n - 1 = 80 - 1 = 79
t = (x - µ)/(s/√n)
Where
x = sample mean = 15.6
µ = population mean = 17
s = samples standard deviation = 4.5
t = (15.6 - 17)/(4.5/√80) = - 2.78
We would determine the p value using the t test calculator. It becomes
p = 0.0034
Since alpha, 0.05 > than the p value, 0.0034, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that students at her college study less than 17 hours per week.
We reject the null hypothesis. The sample data showed significant evidence that students at her college study less than 17 hours per week.
What are null hypotheses and alternative hypotheses?
In null hypotheses, there is no relationship between the two phenomena under the assumption that it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.
Mean (μ) = 15.6
Standard deviation (σ) = 4.5
Sample size (n) = 80
For the null hypotheses
μ = 17
For the alternative hypotheses
μ < 17
The degree of freedom will be
→ n - 1 = 80 - 1 = 79
Then the left tailed test will be
[tex]\rm t = \dfrac{\mu - \bar{x}}{s/\sqrt{n}}\\\\\\t = \dfrac{15.6 - 17 }{4.5 / \sqrt{80}}\\\\\\t = -2.78[/tex]
The p-value using the t-test calculator will be
p = 0.0034
Since, α > 0.05 then the p-value = 0.0034. Then we reject the null hypothesis. Therefore, at a 5% level of significance, the sample data showed significant evidence that students at her college study less than 17 hours per week.
More about the null hypotheses and alternative hypotheses link is given below.
https://brainly.com/question/9504281
