(a) Two-tailed p-value: 2P(Z > |z|) = 0.1615133
It is very difficult to calculate p-value manually. The most commonly employed way of doing this is to utilize a z-score table. In a z-table, the zone under the probability density function is presented for each value of the z-score.
It is also possible to employ an integral to determine the area under the curve. The standard normal distribution function that is used to do this is as follows:
φ(z) = (1 / √2π) × e -z2/2
Where: −∞ < z < ∞, e is the base of the natural logarithm (2.718282), π is the constant (3.1415926).
b)
P-value >
Fail to Reject Null Hypothesis
(10 Points) A 2-tailed z-test was performed. The test statistic was- (5) What is the p-value for ...
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...
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The value obtained for the test statistic, z, in a one-mean 2-test is given. Whether the test is two tailed, left tailed, or right tailed is also specified. For parts (a) and (b). 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. a. The test statistic in a two-tailed test is z= -1.42. The P-value is (Round to three cocimal places as...
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a two tailed test is performed at a 5% level of significance. THE value is determined to be 0.04 the null hypothesis a) has been designed incorrect 2) should not be rejected 3)may or may not be rejected depending on sample size 4. should be rejected
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Question 10 In conducting a hypothesis test with we find a test statistic with a p-value of 0.0389. Let the significance level be .010. The researcher would have: found another significance level rejected the null hypothesis not been able to draw a conclusion about the null hypothesis not rejected the null hypothesis
4) Given the following information what is the p-value of the test? Using the p-value method would you reject or not reject the null hypothesis at a significance level of 0.05? (8 pts) a) The test statistic for a right-tailed test is z = 0.55 b) The test statistic for a two-tailed test is z = 1.95 c) The test statistic in a left-tailed test is z = -1.72. d) The test statistic for a two-tailed test is z...
The test statistic in a two dash tailedtwo-tailed test is zequals=1.79 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 (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...
For a one-tailed hypothesis test, the computed test statistic is z= 2.65. What is the p-value?