Question

A study by Consumer Reports showed that 64% of supermarket shoppers believesupermarket brands to be as good as national brands. A manager of a local storebelieves that the percentage is higher in his store. She asks a random sample of145 shoppers whether they believed that supermarket brand ketchup was as goodas the national brand ketchup. Out of the 145 shoppers asked, 89 believed the supermarket brand was as good as the national brand. Use α= 0.05 significancelevel and conduct a hypothesis test.

Answer 1

`z=\frac{0.614 -0.64}{\sqrt{(0.64(1-0.64))/(145)}}=-0.652`

The p value would be given by:

`p_v =P(z<-0.654)=0.257`

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion is significantly less than 0.64 at 5% of significance

Explanation:

Information given

n=145 represent the random sample taken

X=89 represent the number of people who believed the supermarket brand was as good as the national brand

`\hat p=(89)/(145)=0.614` estimated proportion of people who believed the supermarket brand was as good as the national brand

`p_o=0.64` is the value to test

`\alpha=0.05` represent the significance level

z would represent the statistic

`p_v` represent the p value

Hypothesis to test

We want to test if the true proportion is less than 0.64, the system of hypothesis are.:

Null hypothesis:
`p \geq 0.64`

Alternative hypothesis:
`p < 0.64`

The statistic is given by:

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

Replacing the info we got:

`z=\frac{0.614 -0.64}{\sqrt{(0.64(1-0.64))/(145)}}=-0.652`

The p value would be given by:

`p_v =P(z<-0.654)=0.257`

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion is significantly less than 0.64 at 5% of significance