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.)
solution:-
given that
test statistic t = 3.075
and n = 11
degree of freedom df = n - 1 = 11 - 1 = 10
the P-value for a two tailed test is
=> P-value = 0.0117
Use technology to find the P-value for a two-tailed test with nequals11 and test statistic t...
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 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 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 two-tailed test withn equals=15and test statistic t equals negative t=−2.143.
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 a right-tailed test about a mean with n equals=11 and test statistic t equals 2.258
The test statistic in a two-tailed test is z=0.34 Determine the P-value and decide whether, at the 10% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. LOADING... Click here to view a partial table of areas under the standard normal curve. The P-value is nothing. (Round to four decimal places as needed.) This P-value ▼ sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis because it...
Find the P-value for the indicated hypothesis test with the given standardized test statistic, z. Decide whether to reject Upper H 0H0 for the given level of significance alphaα. Two-tailed test with test statistic zequals=negative 1.86−1.86 and alphaαequals=0.060.06 P-valueequals= nothing (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.)