Question

Consider the following hypotheses: Ho: p2 0.3!5 HA: p<0.35 Compute the p-value based on the following sample information. (You may find it useful to reference the appropriate table: z table or ttable) (Round "z" valueto 2 decimal places. Round intermediate calculations to at least 4 decimal places and final answers to 4 decimal places.) p-value a. x- 30; n130 b.x-85; n-327 С. р- 0.33; n 64 d. p-0.33; n 448

Answer 1

Solution :-

This is the left tailed test .

a) $\hat p$ = x / n = 30/130 = 0.2308

P₀ = 0.35

1 - P₀ = 1-0.35 = 0.65

Test statistic = z

= $\hat p$ - P₀ / [$\sqrt$P_{0 *} (1 - P₀ ) / n]

=0.2308-0.35 / [$\sqrt$0.35(1-0.35) /130 ]

= -2.85

P(z <-2.85 ) = 0.0022

P-value = 0.0022

b)

$\hat p$ = x / n = 85/327= 0.2599

P₀ = 0.35

1 - P₀ = 1-0.35 = 0.65

Test statistic = z

= $\hat p$ - P₀ / [$\sqrt$P_{0 *} (1 - P₀ ) / n]

=0.2599-0.35 / [$\sqrt$0.35(1-0.35) /327 ]

= --3.41

P(z <-3.41 ) = 0.0003

P-value = 0.0003

c) n = 64

$\hat p$ = 0.33

P₀ = 0.35

1 - P₀ = 1-0.35 = 0.65

Test statistic = z

= $\hat p$ - P₀ / [$\sqrt$P_{0 *} (1 - P₀ ) / n]

=0.33 - 0.35 / [$\sqrt$0.35(1-0.35) /64 ]

= -0.34

P(z <-0.34 ) = 0.3669

P-value = 0.3669

d ) n =448

$\hat p$ = 0.33

P₀ = 0.35

1 - P₀ = 1-0.35 = 0.65

Test statistic = z

= $\hat p$ - P₀ / [$\sqrt$P_{0 *} (1 - P₀ ) / n]

=0.33 - 0.35 / [$\sqrt$0.35(1-0.35) /448 ]

= -0.91

P(z <-0.91 ) = 0.1814

P-value = 0.1814