Question

se technology to find the​ P-value for the hypothesis test described below.

The claim is that for a smartphone​ carrier's data speeds at​ airports, the mean is

muμequals=16.00

Mbps. The sample size is

nequals=14

and the test statistic is

tequals=negative −2.404.

​P-valueequals=nothing

​(Round to three decimal places as​ needed.)

se technology to find the​ P-value for the hypothesis test described below.

The claim is that for a smartphone​ carrier's data speeds at​ airports, the mean is

muμequals=16.0016.00

Mbps. The sample size is

nequals=1414

and the test statistic is

tequals=negative 2.404−2.404.

​P-valueequals=nothing

​(Round to three decimal places as​ needed.)

se technology to find the​ P-value for the hypothesis test described below.

The claim is that for a smartphone​ carrier's data speeds at​ airports, the mean is

muμequals=16.0016.00

Mbps. The sample size is

nequals=1414

and the test statistic is

tequals=negative 2.404−2.404.

​P-valueequals=nothing

​(Round to three decimal places as​ needed.)

se technology to find the​ P-value for the hypothesis test described below.

The claim is that for a smartphone​ carrier's data speeds at​ airports, the mean is

muμequals=16.00

Mbps. The sample size is

nequals=14

and the test statistic is

tequals=negative −2.404.

​P-valueequals=nothing

​(Round to three decimal places as​ needed.)

Answer 1

Solution :

The null and alternative hypotheses are as follows :

Mbps

Mbps

The value of the test statistic is, t = -2.404

Sample size (n) = 14

Degrees of freedom = (n - 1) = (14 - 1) = 13

Since, the test is two-tailed test, therefore we shall obtain the two-tailed p-value for the test statistic. The two-tailed p-value is given as follows :

p-value = 2.P(T > |t|)

We have, |t| = 2.404

p-value = 2.P(T > 2.404)

Using R software we get, 2.P(T > 2.404) = 0.032

Hence, p-value = 0.032

The p-value is 0.032.