Answer :
### Question 1: Testing [tex]\( H_0: p = 0.5 \)[/tex] against [tex]\( H_a: p > 0.5 \)[/tex]
Given:
- Sample size [tex]\( n = 120 \)[/tex]
- Sample proportion [tex]\( \hat{p} = 0.76 \)[/tex]
- Null hypothesis proportion [tex]\( p_0 = 0.5 \)[/tex]
- Significance level [tex]\( \alpha = 0.01 \)[/tex]
- Critical [tex]\( z \)[/tex] score for [tex]\( \alpha = 0.01 \)[/tex] is 2.33
The test statistic [tex]\( z \)[/tex] is calculated using the formula:
[tex]\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \][/tex]
Plugging in the values:
[tex]\[ z = \frac{0.76 - 0.5}{\sqrt{\frac{0.5 \times 0.5}{120}}} = 5.696 \][/tex]
So,
test statistic [tex]\( z = 5.696 \)[/tex].
Since [tex]\( z = 5.696 \)[/tex] is greater than the critical [tex]\( z \)[/tex] score of 2.33, we reject the null hypothesis.
Conclusion:
A. We can reject the null hypothesis that [tex]\( p = 0.5 \)[/tex] and accept that [tex]\( p > 0.5 \)[/tex].
### Question 2: Testing [tex]\( H_0: p = 0.55 \)[/tex] against [tex]\( H_a: p < 0.55 \)[/tex]
Given:
- Sample size [tex]\( n = 120 \)[/tex]
- Sample proportion [tex]\( \hat{p} = 0.76 \)[/tex]
- Null hypothesis proportion [tex]\( p_0 = 0.55 \)[/tex]
- Significance level [tex]\( \alpha = 0.05 \)[/tex]
- Critical [tex]\( z \)[/tex] score for [tex]\( \alpha = 0.05 \)[/tex] is -1.645
The test statistic [tex]\( z \)[/tex] is calculated using the formula:
[tex]\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \][/tex]
Plugging in the values:
[tex]\[ z = \frac{0.76 - 0.55}{\sqrt{\frac{0.55 \times 0.45}{120}}} = 4.624 \][/tex]
So,
test statistic [tex]\( z = 4.624 \)[/tex].
Since [tex]\( z = 4.624 \)[/tex] is not lesser than the critical [tex]\( z \)[/tex] score of -1.645, we do not reject the null hypothesis.
Conclusion:
A. There is not sufficient evidence to reject the null hypothesis that [tex]\( p = 0.55 \)[/tex].
### Question 3: Testing [tex]\( H_0: p = 0.55 \)[/tex] against [tex]\( H_a: p \neq 0.55 \)[/tex]
Given:
- Sample size [tex]\( n = 120 \)[/tex]
- Sample proportion [tex]\( \hat{p} = 0.76 \)[/tex]
- Null hypothesis proportion [tex]\( p_0 = 0.55 \)[/tex]
- Significance level [tex]\( \alpha = 0.05 \)[/tex]
- Critical [tex]\( z \)[/tex] score for two-tailed test with [tex]\( \alpha = 0.05 \)[/tex] is 1.96
Since we already calculated the test statistic for [tex]\( H_0: p = 0.55 \)[/tex] against [tex]\( H_a: p < 0.55 \)[/tex], which is:
[tex]\[ z = 4.624 \][/tex]
The test statistic [tex]\( z = 4.624 \)[/tex] remains the same.
Since [tex]\( |z| = 4.624 \)[/tex] is greater than the critical [tex]\( z \)[/tex] score of 1.96, we reject the null hypothesis.
Conclusion:
B. We can reject the null hypothesis that [tex]\( p = 0.55 \)[/tex] and accept that [tex]\( p \neq 0.55 \)[/tex].
Given:
- Sample size [tex]\( n = 120 \)[/tex]
- Sample proportion [tex]\( \hat{p} = 0.76 \)[/tex]
- Null hypothesis proportion [tex]\( p_0 = 0.5 \)[/tex]
- Significance level [tex]\( \alpha = 0.01 \)[/tex]
- Critical [tex]\( z \)[/tex] score for [tex]\( \alpha = 0.01 \)[/tex] is 2.33
The test statistic [tex]\( z \)[/tex] is calculated using the formula:
[tex]\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \][/tex]
Plugging in the values:
[tex]\[ z = \frac{0.76 - 0.5}{\sqrt{\frac{0.5 \times 0.5}{120}}} = 5.696 \][/tex]
So,
test statistic [tex]\( z = 5.696 \)[/tex].
Since [tex]\( z = 5.696 \)[/tex] is greater than the critical [tex]\( z \)[/tex] score of 2.33, we reject the null hypothesis.
Conclusion:
A. We can reject the null hypothesis that [tex]\( p = 0.5 \)[/tex] and accept that [tex]\( p > 0.5 \)[/tex].
### Question 2: Testing [tex]\( H_0: p = 0.55 \)[/tex] against [tex]\( H_a: p < 0.55 \)[/tex]
Given:
- Sample size [tex]\( n = 120 \)[/tex]
- Sample proportion [tex]\( \hat{p} = 0.76 \)[/tex]
- Null hypothesis proportion [tex]\( p_0 = 0.55 \)[/tex]
- Significance level [tex]\( \alpha = 0.05 \)[/tex]
- Critical [tex]\( z \)[/tex] score for [tex]\( \alpha = 0.05 \)[/tex] is -1.645
The test statistic [tex]\( z \)[/tex] is calculated using the formula:
[tex]\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \][/tex]
Plugging in the values:
[tex]\[ z = \frac{0.76 - 0.55}{\sqrt{\frac{0.55 \times 0.45}{120}}} = 4.624 \][/tex]
So,
test statistic [tex]\( z = 4.624 \)[/tex].
Since [tex]\( z = 4.624 \)[/tex] is not lesser than the critical [tex]\( z \)[/tex] score of -1.645, we do not reject the null hypothesis.
Conclusion:
A. There is not sufficient evidence to reject the null hypothesis that [tex]\( p = 0.55 \)[/tex].
### Question 3: Testing [tex]\( H_0: p = 0.55 \)[/tex] against [tex]\( H_a: p \neq 0.55 \)[/tex]
Given:
- Sample size [tex]\( n = 120 \)[/tex]
- Sample proportion [tex]\( \hat{p} = 0.76 \)[/tex]
- Null hypothesis proportion [tex]\( p_0 = 0.55 \)[/tex]
- Significance level [tex]\( \alpha = 0.05 \)[/tex]
- Critical [tex]\( z \)[/tex] score for two-tailed test with [tex]\( \alpha = 0.05 \)[/tex] is 1.96
Since we already calculated the test statistic for [tex]\( H_0: p = 0.55 \)[/tex] against [tex]\( H_a: p < 0.55 \)[/tex], which is:
[tex]\[ z = 4.624 \][/tex]
The test statistic [tex]\( z = 4.624 \)[/tex] remains the same.
Since [tex]\( |z| = 4.624 \)[/tex] is greater than the critical [tex]\( z \)[/tex] score of 1.96, we reject the null hypothesis.
Conclusion:
B. We can reject the null hypothesis that [tex]\( p = 0.55 \)[/tex] and accept that [tex]\( p \neq 0.55 \)[/tex].
