Question

Consider the following hypothesis test:

H₀: p ≥ 0.75 

Hα: p<0.75

A sample of 400 items was selected. Compute the p-value and state your conclusion for each of the following sample results. Use α=.05.

Round your answers to four decimal places.

a. p̅=0.69

b. p̅=0.72

c. p̅=0.7

d. p̅=0.79

Answer 1

This is the left tailed test .

The null and alternative hypothesis is

H₀ : p = 0.75

H_a : p < 0.75

n =400

a ) $\hat p$ = 0.61

P₀ = 0.75

1 - P₀ = 1 - 0.75 = 0.25

Test statistic = z

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

= 0.61 - 0.75/ [$\sqrt$0.75 * 0.25 / 400]

= −6.466

Test statistic = z = −6.47

P-value = 0

$\alpha$ = 0.05

P-value < $\alpha$

0 < 0.05

Reject the null hypothesis .

There is sufficient evidence to suggest that < 0.75

b ) $\hat p$ = 0.72

P₀ = 0.75

1 - P₀ = 1 - 0.75 = 0.25

Test statistic = z

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

= 0.72 - 0.75/ [$\sqrt$0.75 * 0.25 / 400]

= −1.386

Test statistic = z = −1.39

P-value = 0.0829

$\alpha$ = 0.05

P-value ≥$\alpha$

0.0829 ≥ 0.05

Do not reject the null hypothesis .

There is sufficient evidence to suggest that < 0.75

c ) $\hat p$ = 0.70

P₀ = 0.75

1 - P₀ = 1 - 0.75 = 0.25

Test statistic = z

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

= 0.70 - 0.75/ [$\sqrt$0.75 * 0.25 / 400]

= −2.309

Test statistic = z = −2.31

P-value = 0.0105

$\alpha$ = 0.05

P-value < $\alpha$

0.0105< 0.05

Reject the null hypothesis .

There is sufficient evidence to suggest that < 0.75

d ) $\hat p$ = 0.79

P₀ = 0.75

1 - P₀ = 1 - 0.75 = 0.25

Test statistic = z

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

= 0.79 - 0.75/ [$\sqrt$0.75 * 0.25 / 400]

= 1.848

Test statistic = z = 1.85

P-value = 0.9677

$\alpha$ = 0.05

P-value ≥ $\alpha$

0.9677 ≥0.05

Do not reject the null hypothesis

There is sufficient evidence to suggest that < 0.75

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