Question

your manager believes that 43% of the company's orders come from first-time customers. A random sample of 98 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.33 and 0.46?

Answer 1

Answer: 0.7030

Explanation:

Given : The population proportion for company's orders come from first-time customers : p=0.43

Sample size : n= 98

The test statistic for population proportion:-

`z=\frac{\hat{p}-p}{\sqrt{(p(1-p))/(n)}}`

For ,
`\hat{p}=0.33`

`z=\frac{0.33-0.43}{\sqrt{(0.43(1-0.43))/(98)}}\approx-2.00`

For ,
`\hat{p}=0.46`

`z=\frac{0.46-0.43}{\sqrt{(0.43(1-0.43))/(98)}}\approx0.60`

`\text{The p-value =}P(-2.00<z<0.60)=P(z<0.60)-P(z<-2.00)\\\\=0.7257469-0.0227501=0.7029968\approx0.7030`

Hence, the probability that the sample proportion is between 0.33 and 0.46 is 0.7030.