To conduct this statistical test, let's start by identifying the correct hypotheses and test type.
Hypotheses
Since we want to test whether the proportion of people who are confident is smaller than 90%, the appropriate null and alternative hypotheses are:
B. H_0 : p ≥ 0.9
H_1 : p < 0.9
Test Type
Since we are testing if the proportion is less than (i.e., we are evaluating if it has decreased from 90%), the test is:
B. left-tailed
Test Statistic Calculation
We will use the formula for the test statistic for proportions:
\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]
Where:
- \(\hat{p} = 0.81\) (sample proportion)
- \(p_0 = 0.9\) (hypothesized population proportion)
- \(n = 200\) (sample size)
Plugging in the values:
-
Calculate the standard error: \[ SE = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.9(1 - 0.9)}{200}} = \sqrt{\frac{0.9 \times 0.1}{200}} = \sqrt{\frac{0.09}{200}} \approx \sqrt{0.00045} \approx 0.0212 \]
-
Now, calculate the z statistic: \[ z = \frac{0.81 - 0.9}{0.0212} \approx \frac{-0.09}{0.0212} \approx -4.245 \]
So, the test statistic is approximately:
Test Statistic: -4.245 (to 3 decimals: -4.245)
P-value Calculation
We will find the p-value corresponding to the calculated z-value in a standard normal distribution. The p-value for a left-tailed test can be found using a standard normal distribution table or calculation.
Looking up \(z = -4.245\), or using a calculator:
\[ p\text{-value} \approx 0.000014 (to 4 decimals: 0.0000) \]
Conclusion
Since the p-value is significantly lower than the significance level \(\alpha = 0.05\), we have sufficient evidence to reject the null hypothesis.
Thus, based on the calculated values:
Final Answers:
Test statistic is: -4.245
p-value is: 0.0000
Based on this we: B. Reject the null hypothesis
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