Question

twenty years ago, a very famous psychologist specializing in marriage counseling authored a book detailing the way in which she believed spouses should communicate. she is now interested in the proportion of all couples who bought her book who stayed together. for a random sample of 275 couples who bought her book, she found that 225 of them stayed together. based on this, compute a 90% confidence interval for the proportion of all couples who bought her book who stayed together. then find the lower limit and upper limit of the 90% confidence interval. carry your intermediate computations to at least three decimal places. round your answers to two decimal places. (if necessary, consult a list of formulas.) lower limit: upper limit:

Explanation:

Step1: Calculate the sample proportion

The sample proportion $\hat{p}$ is given by $\hat{p}=\frac{x}{n}$, where $x = 225$ (number of successes) and $n=275$ (sample size).
$\hat{p}=\frac{225}{275}\approx0.818$

Step2: Find the critical value

For a $90\%$ confidence interval, the significance level $\alpha=1 - 0.90=0.10$, and $\alpha/2=0.05$. The critical value $z_{\alpha/2}$ is such that $P(Z>z_{\alpha/2}) = 0.05$. From the standard normal table, $z_{\alpha/2}=z_{0.05}\approx1.645$

Step3: Calculate the standard error

The standard error $SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.818$ and $n = 275$:
$SE=\sqrt{\frac{0.818\times(1 - 0.818)}{275}}=\sqrt{\frac{0.818\times0.182}{275}}\approx\sqrt{\frac{0.149}{275}}\approx\sqrt{0.000542}\approx0.023$

Step4: Calculate the confidence interval

The confidence interval for a proportion is $\hat{p}\pm z_{\alpha/2}\times SE$.
The lower limit is $\hat{p}-z_{\alpha/2}\times SE=0.818-1.645\times0.023=0.818 - 0.038=0.78$
The upper limit is $\hat{p}+z_{\alpha/2}\times SE=0.818+1.645\times0.023=0.818 + 0.038=0.86$

Answer:

Lower limit: $0.78$
Upper limit: $0.86$