Respuesta :
Answer:
The teenagers spent 4.5 hours per week, on average, on the phone.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 4.5
Sample mean, [tex]\bar{x}[/tex] = 4.75
Sample size, n = 15
Alpha, α = 0.05
Sample standard deviation, s = 2
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 4.5\text{ hours per week}\\H_A: \mu > 4.5\text{ hours per week}[/tex]
We use one-tailed(right) t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{4.75 - 4.5}{\frac{2}{\sqrt{15}} } = 0.4841[/tex]
Now,
[tex]t_{critical} \text{ at 0.05 level of significance, 14 degree of freedom } = 1.76[/tex]
Since,
[tex]t_{stat} < t_{critical}[/tex]
We accept the null hypothesis. Thus, the teenagers spent 4.5 hours per week, on average, on the phone. The sample contradicted the organization's claim.
Answer:
There is no evidence that the population mean is greater than [tex]$4.5$[/tex] hours.
Step-by-step explanation:
Given:
[tex]$\begin{array}{ll}\mu=4.5 & \text { (Population mean) } \\ n=15 & \text { (Sample size) } \\ \bar{x}=4.75 & \text { (Sample mean) } \\ s=2 & \text { (Sample standard deviation) }\end{array}$[/tex]
Step 1 :
Null hypothesis:
[tex]$H_{0}: \mu \leq 4.5$[/tex]
Alternative hypothesis: [tex]$H_{1}: \mu>4.5 \quad$[/tex]
Since the alternative hypothesis has a greater than symbol, it is a right-tailed hypothesis test.
Significance level is [tex]$\alpha=0.05$[/tex]
Step 2:
The Test statistic of the organization
[tex]$t_{t e s t}=\frac{\bar{x}-\mu}{s / \sqrt{n}}$[/tex]
[tex]$=\frac{4.75-4.5}{2 / \sqrt{15}}$[/tex]
[tex]$=\frac{0.25}{2 / 3.87298335}$[/tex]
[tex]$=\frac{0.25}{0.516397779}$[/tex]
[tex]$\approx 0.484$[/tex]
The p-value for a right-tailed test is given by:
[tex]$P-\text { value }=P\left(t>\left|t_{\text {test }}\right|\right)[/tex]
[tex]\Rightarrow P-\text { value }=P(t>|0.484|)$[/tex]
Step 3:
Now,
[tex]$t_{\text {critical }}$[/tex] at [tex]$0.05$[/tex] level of significance, [tex]14[/tex] degree of freedom [tex]$=1.76$[/tex]
Since,
[tex]$t_{\text {stat }}<t_{\text {critical }}$[/tex]
so, accept the null hypothesis. Thus, the teenagers spent [tex]$4.5$[/tex] hours per week, on average, on the phone. The sample contradicted the organization's claim.
To learn more information, refer:
https://brainly.com/question/22850174
