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Perform a one-sample proportion test with continuity correction for the data, given the null probability is 0.5 and the alternative hypothesis is that the true proportion is not equal to 0.5.
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Step-by-step solution
5 stepsIdentify the type of test performed.
The command
prop.test(16,50)indicates that a one-sample proportion test was conducted. This test is used to compare an observed proportion to a hypothesized population proportion.State the null and alternative hypotheses.
The output explicitly states:
- Null Hypothesis (): The true population proportion () is equal to 0.5.
- Alternative Hypothesis (): The true population proportion () is not equal to 0.5. This is a two-sided test.
Identify the test statistic and its p-value.
The output provides the chi-squared test statistic and its corresponding p-value:
- Test statistic:
- Degrees of freedom:
- P-value:
Interpret the results based on the p-value.
Since the p-value (0.01621) is less than the common significance level of 0.05, we reject the null hypothesis. This means there is statistically significant evidence to conclude that the true population proportion is different from 0.5.
State the 95% confidence interval for the true proportion.
The output provides the 95% confidence interval:
- Lower bound: 0.1992781
- Upper bound: 0.4683118 This means we are 95% confident that the true population proportion lies between approximately 0.1993 and 0.4683.
Summary of results:
- The test performed is a one-sample proportions test.
- The null hypothesis is that the true proportion is 0.5.
- The alternative hypothesis is that the true proportion is not 0.5.
- The chi-squared test statistic is with .
- The p-value is .
- Since , we reject the null hypothesis, concluding that the true proportion is significantly different from 0.5.
- The 95% confidence interval for the true proportion is .
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