Question

The display provided from technology available below results from using data for a smartphone​ carrier's data speeds at airports to test the claim that they are from a population having a mean less than 4.004.00 Mbps. Conduct the hypothesis test using these results. Use a 0.050.05 significance level. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim. LOADING... Click the icon to view the display from technology. What are the null and alternative​ hypotheses? A. Upper H 0H0​: muμequals=4.004.00 Mbps

Answer 1

Answer:

Null hypothesis:$\mu \geq 4.0$    

Alternative hypothesis:$\mu < 4.00$    

$t=\frac{3.48-4.00}{\frac{1.150075}{\sqrt{45}}}=-3.033077$    

$df=n-1=45-1=44$

Since is a left-sided test the p value would be:    

$p_v =P(t_{(44)}$    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly less than 4.00 at 5% of significance.  

Step-by-step explanation:

Data given and notation    

$\bar X=3.48$ represent the sample mean

$s=1.150075$ represent the sample standard deviation    

$n=45$ sample size    

$\mu_o =4.00$ represent the value that we want to test  

$\alpha=0.05$ represent the significance level for the hypothesis test.    

z would represent the statistic (variable of interest)    

$p_v$ represent the p value for the test (variable of interest)    

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is less than 4.00 :    

Null hypothesis:$\mu \geq 4.0$    

Alternative hypothesis:$\mu < 4.00$    

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

$t=\frac{3.48-4.00}{\frac{1.150075}{\sqrt{45}}}=-3.033077$    

P-value    

First we need to calculate the degrees of freedom given by:

$df=n-1=45-1=44$

Since is a left-sided test the p value would be:    

$p_v =P(t_{(44)}$    

Conclusion    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly less than 4.00 at 5% of significance.