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Sagot :
Let's go through the problem step-by-step:
a) Determine the null and alternative hypotheses.
- The null hypothesis ([tex]\(H_0\)[/tex]) represents the status quo or the current accepted value. In this case, it states that 71% of college students work.
[tex]\[ H_0: p = 0.71 \][/tex]
- The alternative hypothesis ([tex]\(H_a\)[/tex]) represents the claim being tested. The researcher thinks the percentage of college students who work has changed. So, the alternative hypothesis is:
[tex]\[ H_a: p \neq 0.71 \][/tex]
b) Determine the test statistic. Round to two decimals.
- To find the test statistic, we use the formula for the z-score in the context of proportions:
[tex]\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \][/tex]
where:
[tex]\(\hat{p}\)[/tex] = sample proportion
[tex]\(p_0\)[/tex] = population proportion
[tex]\(n\)[/tex] = sample size
- Given:
[tex]\[ \hat{p} = \frac{91}{110} = 0.8273 \][/tex]
[tex]\[ p_0 = 0.71 \][/tex]
[tex]\[ n = 110 \][/tex]
- Calculating the standard error:
[tex]\[ SE = \sqrt{\frac{0.71 \times (1 - 0.71)}{110}} \approx 0.0425 \][/tex]
- Calculating the z-score:
[tex]\[ z = \frac{0.8273 - 0.71}{0.0425} \approx 2.76 \][/tex]
c) Find the p-value. Round to 4 decimals.
- To find the p-value associated with the calculated z-score for a two-tailed test:
[tex]\[ p\text{-value} = 2 \times P(Z > 2.76) \][/tex]
- Using standard normal distribution tables or computational tools, we find:
[tex]\[ p\text{-value} \approx 0.0057 \][/tex]
d) Make a decision.
- The significance level ([tex]\(\alpha\)[/tex]) is 0.01. We compare the p-value to the significance level:
[tex]\[ 0.0057 < 0.01 \][/tex]
- Since the p-value is less than the significance level, we reject the null hypothesis.
e) Write the conclusion.
- Given that we rejected the null hypothesis, we conclude that:
[tex]\[ \text{There is sufficient evidence to support the claim that the percentage of college students who work is different than 71\%.} \][/tex]
Summary:
a) [tex]\(H_0: p = 0.71\)[/tex], [tex]\(H_a: p \neq 0.71\)[/tex]
b) [tex]\(z = 2.76\)[/tex]
c) [tex]\(p\text{-value} = 0.0057\)[/tex]
d) Reject the null hypothesis
e) There is sufficient evidence to support the claim that the percentage of college students who work is different than 71\%.
a) Determine the null and alternative hypotheses.
- The null hypothesis ([tex]\(H_0\)[/tex]) represents the status quo or the current accepted value. In this case, it states that 71% of college students work.
[tex]\[ H_0: p = 0.71 \][/tex]
- The alternative hypothesis ([tex]\(H_a\)[/tex]) represents the claim being tested. The researcher thinks the percentage of college students who work has changed. So, the alternative hypothesis is:
[tex]\[ H_a: p \neq 0.71 \][/tex]
b) Determine the test statistic. Round to two decimals.
- To find the test statistic, we use the formula for the z-score in the context of proportions:
[tex]\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \][/tex]
where:
[tex]\(\hat{p}\)[/tex] = sample proportion
[tex]\(p_0\)[/tex] = population proportion
[tex]\(n\)[/tex] = sample size
- Given:
[tex]\[ \hat{p} = \frac{91}{110} = 0.8273 \][/tex]
[tex]\[ p_0 = 0.71 \][/tex]
[tex]\[ n = 110 \][/tex]
- Calculating the standard error:
[tex]\[ SE = \sqrt{\frac{0.71 \times (1 - 0.71)}{110}} \approx 0.0425 \][/tex]
- Calculating the z-score:
[tex]\[ z = \frac{0.8273 - 0.71}{0.0425} \approx 2.76 \][/tex]
c) Find the p-value. Round to 4 decimals.
- To find the p-value associated with the calculated z-score for a two-tailed test:
[tex]\[ p\text{-value} = 2 \times P(Z > 2.76) \][/tex]
- Using standard normal distribution tables or computational tools, we find:
[tex]\[ p\text{-value} \approx 0.0057 \][/tex]
d) Make a decision.
- The significance level ([tex]\(\alpha\)[/tex]) is 0.01. We compare the p-value to the significance level:
[tex]\[ 0.0057 < 0.01 \][/tex]
- Since the p-value is less than the significance level, we reject the null hypothesis.
e) Write the conclusion.
- Given that we rejected the null hypothesis, we conclude that:
[tex]\[ \text{There is sufficient evidence to support the claim that the percentage of college students who work is different than 71\%.} \][/tex]
Summary:
a) [tex]\(H_0: p = 0.71\)[/tex], [tex]\(H_a: p \neq 0.71\)[/tex]
b) [tex]\(z = 2.76\)[/tex]
c) [tex]\(p\text{-value} = 0.0057\)[/tex]
d) Reject the null hypothesis
e) There is sufficient evidence to support the claim that the percentage of college students who work is different than 71\%.
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