Question

(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.

Answer 1

(a) Two-tailed p-value: 2P(Z > |z|) = 0.1615133

Two-tailed p-value: 2P(Z > IzI) 0.1615133 -6 5 -3-2-1 023 4 5 6 The p-value is twice the tail area beyond z

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 > $\alpha$

Fail to Reject Null Hypothesis