Question

a retail manager is analyzing the percentage of customers who make a purchase when they visit a store. based on historical data, the manager believes that a customer makes a purchase 32% of the time. to assess this, 115 customer visits were randomly selected. a. assuming that the managers estimate is correct and that the probability of a customer making a purchase is p = 0.32, compute the corresponding parameters of the sampling distribution and evaluate whether it is appropriate to approximate the sampling distribution with a normal distribution. (round to four decimal places where applicable.) the parameters of the sampling distribution are e(\\(\overline{p}\\)) = and \\(\sigma_{\overline{p}}\\) = it ? appropriate to approximate the sampling distribution with a normal distribution because np = and n(1 - p) = b. calculate the probability that the sample proportion \\(\overline{p}\\) will be between 0.22 and 0.42. (round your answer to four decimal places.) c. calculate the probability that the sample proportion \\(\overline{p}\\) will be between 0.27 and 0.37. (round your answer to four decimal places.)

Explanation:

Step1: Check normal - approximation conditions

We know that \(n = 115\) and \(p=0.32\). Calculate \(np\) and \(n(1 - p)\).

$$np=115\times0.32 = 36.8$$

$$n(1 - p)=115\times(1 - 0.32)=115\times0.68 = 78.2$$

Since \(np\geq10\) and \(n(1 - p)\geq10\), it is appropriate to approximate the sampling distribution with a normal distribution. The mean of the sampling distribution of the sample proportion \(\hat{p}\) is \(E(\hat{p})=p = 0.32\), and the standard deviation is \(\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.32\times(1 - 0.32)}{115}}=\sqrt{\frac{0.32\times0.68}{115}}\approx\sqrt{\frac{0.2176}{115}}\approx\sqrt{0.0019}\approx0.0436\)

Step2: Standard - ize the bounds for part b

For \(\hat{p}_1 = 0.22\), the z - score is \(z_1=\frac{\hat{p}_1 - E(\hat{p})}{\sigma_{\hat{p}}}=\frac{0.22 - 0.32}{0.0436}\approx\frac{- 0.1}{0.0436}\approx - 2.29\)
For \(\hat{p}_2 = 0.42\), the z - score is \(z_2=\frac{\hat{p}_2 - E(\hat{p})}{\sigma_{\hat{p}}}=\frac{0.42 - 0.32}{0.0436}\approx\frac{0.1}{0.0436}\approx2.29\)
\(P(0.22<\hat{p}<0.42)=P(-2.29 < Z < 2.29)=\Phi(2.29)-\Phi(-2.29)\)
Since \(\Phi(-z)=1 - \Phi(z)\), \(P(-2.29 < Z < 2.29)=\Phi(2.29)-(1 - \Phi(2.29)) = 2\Phi(2.29)-1\)
From the standard normal table, \(\Phi(2.29)=0.9890\), so \(P(-2.29 < Z < 2.29)=2\times0.9890 - 1=0.9780\)

Step3: Standard - ize the bounds for part c

For \(\hat{p}_3 = 0.27\), the z - score is \(z_3=\frac{\hat{p}_3 - E(\hat{p})}{\sigma_{\hat{p}}}=\frac{0.27 - 0.32}{0.0436}\approx\frac{-0.05}{0.0436}\approx - 1.15\)
For \(\hat{p}_4 = 0.37\), the z - score is \(z_4=\frac{\hat{p}_4 - E(\hat{p})}{\sigma_{\hat{p}}}=\frac{0.37 - 0.32}{0.0436}\approx\frac{0.05}{0.0436}\approx1.15\)
\(P(0.27<\hat{p}<0.37)=P(-1.15 < Z < 1.15)=\Phi(1.15)-\Phi(-1.15)\)
Since \(\Phi(-z)=1 - \Phi(z)\), \(P(-1.15 < Z < 1.15)=\Phi(1.15)-(1 - \Phi(1.15)) = 2\Phi(1.15)-1\)
From the standard normal table, \(\Phi(1.15)=0.8749\), so \(P(-1.15 < Z < 1.15)=2\times0.8749 - 1 = 0.7498\)

Answer:

a. \(E(\hat{p}) = 0.32\), \(\sigma_{\hat{p}}\approx0.0436\)
b. \(0.9780\)
c. \(0.7498\)