Question

A grocery store manager is interested in testing the claim that banana is the favorite fruit for more than 50% of the adults. The manager conducted a survey on a random sample of 100 adults. The survey showed that 56 adults in the sample chosen banana as his/her favorite fruit. Assume the manager wants to use a 0.05 significance level to test the claim.

Answer 1

Answer:

Claim is False

Step-by-step explanation:

We are given that The manager conducted a survey on a random sample of 100 adults.

The survey showed that 56 adults in the sample chosen banana as his/her favorite fruit.  

Claim :The favorite fruit for more than 50% of the adults.

n =  100

x = 56

We will use one sample proportion test  

$\widehat{p}=\frac{x}{n}$

$\widehat{p}=\frac{56}{100}$

$\widehat{p}=0.56$

The favorite fruit for more than 50% of the adults.

$H_0:p = 0.5\\H_a:p>0.5$

Formula of test statistic =$\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

                                       =$\frac{0.56-0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}$

                                       =$1.2$

Now refer the p value from the z table

p value =0.8849

α =0.05

So, p value >  α

So, we accept the null hypothesis

So,claim is False that  banana is the favorite fruit for more than 50% of the adults.