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

Step-by-step 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:-

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

For , $\hat{p}=0.33$

$z=\dfrac{0.33-0.43}{\sqrt{\dfrac{0.43(1-0.43)}{98}}}\approx-2.00$

For , $\hat{p}=0.46$

$z=\dfrac{0.46-0.43}{\sqrt{\dfrac{0.43(1-0.43)}{98}}}\approx0.60$

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

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