Use technology to find
the P-value for a two-tailed test with
n equals=15
and test statistic t equals negative t=−2.143.
We have given that, test statistic t = -2.143
n = 15
Hence, degrees of freedom = n - 1 = 15-1 = 14
The test is two tailed test.
P-value = 2 * P( T < test statistic)
P-value = 2 * P (T < -2.143)
Using technology that is Excel we use the command,
=TDIST(2.143, 14, 2) we get the value 0.0502.
Hence, P-value = 0.0502
Use technology to find the P-value for a two-tailed test with n equals=15 and test statistic t equals negative t=−2.143.
Use technology to find the P-value for a two-tailed test with nequals11 and test statistic t equals 3.075 . P-valuealmost equals nothing (Round to four decimal places as needed.)
Use technology to find the P-value for a right-tailed test about a mean with n equals=11 and test statistic t equals 2.258
Use technology to find the P-value for a right-tailed test about a mean with n = 9 and test statistic t = 1.451. P-value(Round to four decimal places as needed.)
Use technology to find the P-value for a right-tailed test about a mean with n = 21 and test statistic t = 2.752. P-value (Round to four decimal places as needed.)
use technology to find the P- Value for the a right tailed test about a mean with n=14 and test statistics t=2.825
use technology to find the P- Value for the a right tailed test about a mean with n=14 and test statistics t= - 2.925
question #9
Use technology to find the P-value for a right-tailed test about a mean with n=29 and test statistic 1.946. P-value (Round to four decimal places as needed.)
Use 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 mu equals11.00 Mbps. The sample size is nequals29 and the test statistic is t equals negative 1.989. P-value = (Round to three decimal places as needed.)
Use 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=17.00 Mbps. The sample size is n=14 and the test statistic is t=-1.992 P-value equals=? (Round to three decimal places as needed.)