Question

The claim is that for a smartphone​ carrier's data speeds at​ airports, the mean= 17.00 Mbps. The sample size is n = 20 and the test statistic is t = 1.874

Answer 1

The null hypothesis is that the mean speed of the cable internet connection is three megabits per second or less, while the alternative hypothesis is that the mean speed is greater than three megabits per second.

Step-by-step explanation:

The null hypothesis would be that the mean speed of the cable internet connection is three megabits per second or less, while the alternative hypothesis would state that the mean speed is greater than three megabits per second.

Null hypothesis: The mean data speed of a smartphone carrier at airports is 17.00 Mbps.

Alternative hypothesis: The mean data speed of a smartphone carrier at airports is not 17.00 Mbps.

Significance level: 0.05

Test statistic: t = 1.874

Degrees of freedom: n - 1 = 20 - 1 = 19

p-value: 0.069

Conclusion:

Since the p-value (0.069) is greater than the significance level (0.05), we fail to reject the null hypothesis. There is not enough evidence to conclude that the mean data speed of a smartphone carrier at airports is different from 17.00 Mbps.

Answer 2

The null and alternative hypotheses are:

`H_(0):\mu=17.00`

`H_(a):\mu \\eq 17.00`

We are given that the test statistic is:

`t=1.874`

Also we are given the sample size
`n=20`

Now we have to find the t critical value at 0.05 (assumed) significance level for
`df=n-1=20-1=19`

Using the t distribution table, we have:

`t_(0.05,19)=\pm2.093`

Conclusion: Since the t statistic does not lie outside the t critical values, therefore, we fail to reject the null hypothesis and conclude that there is sufficient evidence to support the claim that a smartphone carrier's mean data speed at airports is 17.00 Mbps.