a recent study estimated that 68% of us adults enjoy reading a book while relaxing. how many adults are…

a recent study estimated that 68% of us adults enjoy reading a book while relaxing. how many adults are required for a random sample to obtain a margin of error of at most 0.05 with 90% confidence? find the z - table here. 236 271 335 578

a recent study estimated that 68% of us adults enjoy reading a book while relaxing. how many adults are required for a random sample to obtain a margin of error of at most 0.05 with 90% confidence? find the z - table here. 236 271 335 578

Answer

Explanation:

Step1: Determine the z - score

For a 90% confidence level, the significance level $\alpha=1 - 0.90 = 0.10$. Then $\alpha/2=0.05$. Looking up in the z - table, the z - score $z_{\alpha/2}=z_{0.05}\approx1.645$.

Step2: Identify the estimated proportion

The estimated proportion $\hat{p}=0.68$, so $1-\hat{p}=1 - 0.68 = 0.32$.

Step3: Use the sample - size formula

The formula for sample size $n$ in estimating a proportion is $n=\frac{z_{\alpha/2}^{2}\hat{p}(1 - \hat{p})}{E^{2}}$, where $E$ is the margin of error. Given $E = 0.05$. Substitute the values: $n=\frac{(1.645)^{2}\times0.68\times0.32}{(0.05)^{2}}$. First, calculate $(1.645)^{2}=2.706025$, $0.68\times0.32 = 0.2176$. Then $(1.645)^{2}\times0.68\times0.32=2.706025\times0.2176\approx0.5888$. And $\frac{0.5888}{(0.05)^{2}}=\frac{0.5888}{0.0025}=235.52$. Since the sample size $n$ must be an integer, we round up to $n = 236$.

Answer:

236