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:
<em>H</em>₀<em>:</em><em> </em>The smartphone carrier's mean data speed at airports is not less than 4.00 Mbps, i.e. <em>μ</em> ≥ 4.00.
<em>H</em>ₐ<em>:</em><em> </em>The smartphone carrier's mean data speed at airports is less than 4.00 Mbps, i.e. <em>μ</em> < 4.00.
Compute the test statistic as follows:
![t=\frac{\bar x-\mu}{s/\sqrt{n}}\\\\=\frac{3.92-4.00}{0.51/\sqrt{93}}\\\\=-1.51](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Cbar%20x-%5Cmu%7D%7Bs%2F%5Csqrt%7Bn%7D%7D%5C%5C%5C%5C%3D%5Cfrac%7B3.92-4.00%7D%7B0.51%2F%5Csqrt%7B93%7D%7D%5C%5C%5C%5C%3D-1.51)
Compute the <em>p</em>-value as follows:
![p-value=P(t_{n-1}](https://tex.z-dn.net/?f=p-value%3DP%28t_%7Bn-1%7D%3C-1.51%29%5C%5C%3DP%28t_%7B92%7D%3C-1.51%29%5C%5C%3D0.066)
*Use a <em>t</em>-table.
*If the desired degrees of freedom is not provided use the next higher value.
The <em>p</em>-value = 0.066 > <em>α</em> = 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.