魅影直播

The null hypothesis, denoted by H₀, is the claim that is initially assumed to be true (the 鈥減rior belief鈥� claim). The alternative hypothesis, denoted by H_a, is the assertion that is contradictory to H₀.

The null hypothesis will be rejected in favour of the alternative hypothesis only if sample evidence suggests that H₀ is false. If the sample does not strongly contradict H₀, we will continue to believe in the plausibility of the null hypothesis. The two possible conclusions from a hypothesis-testing analysis are then reject H0 or fail to reject H₀.

Given that,

\({H_0}:\mu = 5\)

\({H_a}:\mu > 5\)

\(z = 1.42\)

Complement rule:

\(P(\overline A ) = 1 - P(A)\)

The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic z.

Since the alternative hypothesis states , \({H_a}:\mu > 5\), the P-value is the probability to the right of the z-score.

Determine the probability using normal probability table in the appendix, which contain the probability to the left of z-scores(use the complement rule),

\(\begin{array}{l}P = P(Z > 1.42)\\ = 1 - P(Z < 1.42)\\ = 1 - 0.9222\\ = 0.0778\end{array}\)

\(P = 7.78\% \)

Given that,

\({H_0}:\mu = 5\)

\({H_a}:\mu > 5\)

\(z = 0.90\)

Complement rule:

\(P(\overline A ) = 1 - P(A)\)

The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic z.

Since the alternative hypothesis states , \({H_a}:\mu > 5\), the P-value is the probability to the right of the z-score.

Determine the probability using normal probability table in the appendix, which contain the probability to the left of z-scores(use the complement rule),

\(\begin{array}{l}P = P(Z > 0.90)\\ = 1 - P(Z < 0.90)\\ = 1 - 0.8159\\ = 0.1841\end{array}\)

\(P = 18.41\% \)

Given that,

\({H_0}:\mu = 5\)

\({H_a}:\mu > 5\)

\(z = 1.96\)

Complement rule:

\(P(\overline A ) = 1 - P(A)\)

The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic z.

Since the alternative hypothesis states , \({H_a}:\mu > 5\), the P-value is the probability to the right of the z-score.

Determine the probability using normal probability table in the appendix, which contain the probability to the left of z-scores(use the complement rule),

\(\begin{array}{l}P = P(Z > 1.96)\\ = 1 - P(Z < 1.96)\\ = 1 - 0.9750\\ = 0.0250\end{array}\)

\(P = 2.50\% \)

Given that,

\({H_0}:\mu = 5\)

\({H_a}:\mu > 5\)

\(z = 2.48\)

Complement rule:

\(P(\overline A ) = 1 - P(A)\)

The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic z.

Since the alternative hypothesis states , \({H_a}:\mu > 5\), the P-value is the probability to the right of the z-score.

Determine the probability using normal probability table in the appendix, which contain the probability to the left of z-scores(use the complement rule),

\(\begin{array}{l}P = P(Z > 2.48)\\ = 1 - P(Z < 2.48)\\ = 1 - 0.9934\\ = 0.0066\end{array}\)

\(P = 0.66\% \)

Given that,

\({H_0}:\mu = 5\)

\({H_a}:\mu > 5\)

\(z = - 0.11\)

Complement rule:

\(P(\overline A ) = 1 - P(A)\)

The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic z.

Since the alternative hypothesis states , \({H_a}:\mu > 5\), the P-value is the probability to the right of the z-score.

Determine the probability using normal probability table in the appendix, which contain the probability to the left of z-scores(use the complement rule),

\(\begin{array}{l}P = P(Z > - 0.11)\\ = 1 - P(Z < - 0.11)\\ = 1 - 0.4562\\ = 0.5438\end{array}\)

\(P = 54.38\% \)