Answer:
The null hypothesis was failed to be rejected.
Step-by-step explanation:
The complete question is:
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 4.00 Mbps. Conduct the hypothesis test using these results. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. The summary statistics are n=93, x overbar=3.92, s=0.51.
Solution:
The hypothesis is:
H₀: The smartphone carrier's mean data speed at airports is not less than 4.00 Mbps, i.e. μ ≥ 4.00.
Hₐ: The smartphone carrier's mean data speed at airports is less than 4.00 Mbps, i.e. μ < 4.00.
Compute the test statistic as follows:
[tex]t=\frac{\bar x-\mu}{s/\sqrt{n}}\\\\=\frac{3.92-4.00}{0.51/\sqrt{93}}\\\\=-1.51[/tex]
Compute the p-value as follows:
[tex]p-value=P(t_{n-1}<-1.51)\\=P(t_{92}<-1.51)\\=0.066[/tex]
*Use a t-table.
*If the desired degrees of freedom is not provided use the next higher value.
The p-value = 0.066 > α = 0.05
The null hypothesis was failed to be rejected.
Thus, it can be concluded that the smartphone carrier's mean data speed at airports is not less than 4.00 Mbps.
The p-value will be "0.066" and the null hypothesis was failed to be rejected and the test statistic will be "-1.51".
According to the question,
[tex]H_0[/tex] : μ [tex]\geq[/tex] 4.00
[tex]H_a[/tex] : μ < 4.00
The test-statistic will be:
→ t = [tex]\frac{\bar x - \mu}{\frac{s}{\sqrt{n} } }[/tex]
By substituting the values,
= [tex]\frac{3.92 - 4.00}{\frac{0.51}{\sqrt{93} } }[/tex]
= -1.51
hence,
The P-value will be:
= P ([tex]t_{n-1}[/tex] < -1.51)
= P (t₉₂ < -1.51)
= 0.066 > α = 0.05
Thus the answer above is correct.
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