Question

Suppose it is believed that in the population, 65% of americans have less than $2,000 saved for an emergency. a random sample of 50 people is taken, and it is found that 30 out of 50 of them have less than $2,000 saved for an emergency. do we have evidence to suggest that less than 65% of the population has such an amount saved for an emergency? show all steps of statistical testing at 5% level of significance: state your hypotheses using symbols, check assumptions, calculate the test statistic, state the decision and conclusion. show all steps for constructing a 95% confidence interval: state your z-multiplier, calculate the interval, and state your conclusion.

Answer 1

To test whether less than 65% of Americans have less than $2,000 saved for an emergency, a hypothesis test and a confidence interval can be used. The hypothesis test reveals that we do not have evidence to suggest that less than 65% of the population has such an amount saved. The 95% confidence interval is (0.474, 0.726), indicating the range of the true proportion with 95% confidence.

Step-by-step explanation:

To test whether there is evidence to suggest that less than 65% of the population has less than $2,000 saved for an emergency, we can conduct a hypothesis test.

1. State the hypotheses:
   Null hypothesis (H0): Proportion of Americans with less than $2,000 saved for an emergency is equal to 65%
   Alternative hypothesis (Ha): Proportion of Americans with less than $2,000 saved for an emergency is less than 65%
2. Check assumptions:
   This is a binomial distribution problem since we are looking at the proportion of people with less than $2,000 saved. We can assume that the sampling is random and independent, and the sample size is large enough (n=50) for the central limit theorem to apply.
3. Calculate the test statistic:
   We can use the Z-test statistic formula:
   Z = (sample proportion - population proportion) / sqrt((population proportion * (1 - population proportion)) / sample size)
   For this problem, the sample proportion is 30/50 = 0.6, the population proportion is 0.65, and the sample size is 50. Plugging these values into the formula, we get:
   Z = (0.6 - 0.65) / sqrt((0.65 * (1 - 0.65)) / 50) = -0.714
4. State the decision and conclusion:
   At a significance level of 0.05, the critical Z-value for a one-tailed test is -1.645. Since the test statistic (-0.714) is greater than the critical value, we fail to reject the null hypothesis. Therefore, we do not have evidence to suggest that less than 65% of the population has less than $2,000 saved for an emergency.

To construct a 95% confidence interval, we can use the formula:
CI = sample proportion ± z * sqrt((sample proportion * (1 - sample proportion)) / sample size)
For this problem, the sample proportion is 0.6, the sample size is 50, and the z-multiplier for a 95% confidence level is 1.96. Plugging these values into the formula, we get:
CI = 0.6 ± 1.96 * sqrt((0.6 * (1 - 0.6)) / 50) = 0.6 ± 0.126

The 95% confidence interval for the true proportion of Americans with less than $2,000 saved for an emergency is (0.474, 0.726). This means that we can be 95% confident that the true proportion falls within this interval.