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

Claim is False

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}=(x)/(n)`

`\widehat{p}=(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{(p(1-p))/(n)}}`

=
`\frac{0.56-0.5}{\sqrt{(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.