My test statistic for a two-tailed test is 0.4. What is my p-value?
The test Statistics for two tailed test is 0.4
The P-value is given by,
P-value = P(Z 0.4) =
0.6892
using Z table
What is the p-value for a two tailed test when the test statistic is z=0.7? 0.484 0.089 0.242 0.944
What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1.73? (Do not round your answer; compute your answer using a value from the table below.) z−1.8−1.7−1.6−1.5−1.40.000.0360.0450.0550.0670.0810.010.0350.0440.0540.0660.0790.020.0340.0430.0530.0640.0780.030.0340.0420.0520.0630.0760.040.0330.0410.0510.0620.0750.050.0320.0400.0490.0610.0740.060.0310.0390.0480.0590.0720.070.0310.0380.0470.0580.0710.080.0300.0380.0460.0570.0690.090.0290.0370.0460.0560.068
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.)
For a one-tailed hypothesis test, the computed test statistic is z= 2.65. What is the p-value?
(10 Points) A 2-tailed z-test was performed. The test statistic was- (5) What is the p-value for this test? 1.4, a. (5) If the significance level was 10%, would the null hypothesis be accepted or rejected? b. (10 Points) A 2-tailed z-test was performed. The test statistic was- (5) What is the p-value for this test? 1.4, a. (5) If the significance level was 10%, would the null hypothesis be accepted or rejected? b.
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.
The test statistic in a two dash tailed test is z= -2.84. Determine the P-value and decide whether, at the 1% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. *Please show me a way to do it on the calculator (ti-84)*
In a two-tailed test, a statistician got a z test statistic of 1.82. What is the p-value? Multiple Choice .0688 .0301 .0874 .0708
You are performing a two-tailed test with test statistic z = 3.106, find the p-value accurate to 4 decimal places. p-value =
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...