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

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] 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

[tex]\bar X=3.48[/tex] represent the sample mean

[tex]s=1.150075[/tex] represent the sample standard deviation

[tex]n=45[/tex] sample size

[tex]\mu_o =4.00[/tex] represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

[tex]p_v[/tex] 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:[tex]\mu \geq 4.0[/tex]

Alternative hypothesis:[tex]\mu < 4.00[/tex]

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:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (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:

[tex]t=\frac{3.48-4.00}{\frac{1.150075}{\sqrt{45}}}=-3.033077[/tex]

P-value

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

[tex]df=n-1=45-1=44[/tex]

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

[tex]p_v =P(t_{(44)}<-3.033077)=0.002025[/tex]

Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] 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.

ankitprmr2 ankitprmr2 · Answer 2 · 2021-10-17T09:20:22+02:00

[tex]p_v < \alpha[/tex]

so we can conclude that we can 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:

Given :

Sample Mean, [tex]\rm \bar {X}[/tex] = 3.48

Sample Standard Deviation, s = 1.150075

Sample Size, n = 45

The value that we want to test , [tex]\mu_0 = 4[/tex]

Significance level for the hypothesis test, [tex]\alpha = 0.05[/tex]

Calculation :

Let, z be represent the statistic (variable of interest) and [tex]p_v[/tex] represent the p value for the test (variable of interest)

Now, we need to conduct a hypothesis in order to check if the mean is less than 4 :

Null hypothesis - [tex]\mu \geq 4[/tex]

Alternate hypothesis - [tex]\mu < 4[/tex]

Since we don't know the population deviation, is better apply a t test and is given by:

[tex]\rm t = \dfrac{\bar{X}-\mu_0}{\dfrac{s}{\sqrt{n} }}[/tex]

[tex]\rm t = \dfrac{3.48-4}{\dfrac{1.150075}{\sqrt{45} }}[/tex]

Now, calculate the degrees of freedom,

[tex]\rm df = n-1 = 45-1=44[/tex]

Left-sided test the p value would be,

[tex]\rm p_v = P(t_4_4 < -3.033077) = 0.002025[/tex]

[tex]p_v < \alpha[/tex]

so we can conclude that we can reject the null hypothesis, so we can conclude that the mean is significantly less than 4.00 at 5% of significance.

For more information, refer the link given below

https://brainly.com/question/14790912?referrer=searchResults