Question

In: Statistics and Probability

1. A survey of 40 individuals reveals that 19% plan to move within the next year....

1. A survey of 40 individuals reveals that 19% plan to move within the next year. Find a 95% confidence interval for the proportion of people who plan to move next year.

2. In town A, a survey of 54 families finds a mean household size of 2.13, with a standard deviation of 1.0. In town B, a survey of 65 families finds a mean household size of 2.39, with a standard deviation of 1.2. Find an 86% confidence interval around the difference in means.

Solutions

Expert Solution

Answer 1

We need to construct the 95% confidence interval for the population proportion. We have been provided with the following information:
The sample size is N = 40, and the sample proportion is pˉ​​=0.19, and the significance level is α=0.05

Critical Value
Based on the information provided, the significance level is α=0.05, therefore the critical value for this Two-tailed test is Zc​=1.96. This can be found by either using excel or the Z distribution table.

Margin of Error


The confidence interval:

Therefore, based on the data provided, the 95% confidence interval for the population proportion is 0.0684<p<0.3116, which indicates that we are 95% confident that the true population proportion p is contained by the interval (0.0684,0.3116)

Answer 2

Let me know in the comments if anything is not clear. I will reply ASAP! Please do upvote if satisfied!

orchestra answered 2 years ago
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