Question

Recently a case of food poisoning was traced to a particular restaurant chain. The source was identified and corrective actions were taken to make sure that the food poisoning would not reoccur. Despite the response from the restaurant chain, many consumers refused to visit the restaurant for some time after the event. A survey was conducted three months after the food poisoning occurred, with a sample of 319 former customers contacted. 29 indicated that they would not go back to the restaurant because of the potential for food poisoning. Construct a 95 percent confidence interval for the true proportion of the market who still refuse to visit any of the following restaurants in the chain three months after the event.

{.059 0122}

{.090,.091}

{.000, .196}

{.240, .339}

{.118, .244}

Answer 1

Solution :

Given that,

n = 319

x = 29

Point estimate = sample proportion = = x / n = 29/319=0.091

1 - = 1-0.091=0.909

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96} ( Using z table )   

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.091*0.909) / 319)

E = 0.0316

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.091- 0.0316< p < 0.091+0.0316

0.059< p < 0.122

The 95% confidence interval for the population proportion p is : 0.059 , 0.122