Question

A research center claims that at most 18% of U.S. adults’ online food purchases are for snacks. In a random sample of 1,800 U.S. adults, 20.1% say their online food purchases are for snacks. At alpha equals 0.10, is there enough evidence to reject the center’s claim?

                                                                                                                       

1. Identify the null hypothesis, H naught.

1. Identify the alternative hypothesis, H sub a.

1. Find the standardized test statistic, z.     

1. For alpha equals 0.10 , find and describe the rejection region(s).   Make certain to identify the appropriate z-score(s).

1. Determine whether to reject or fail to reject .

1. Interpret this decision in the context of the original claim.

Answer 1

given data are:-

sample size (n) = 1800

sample proportion ($\hat{p}$) = 20.1% = 0.201

hypothesis:-

[null hypothesis]

Hop < 0.18

[ alternative hypothesis ]

H:p> 0.18

the center's claim is the null hypothesis.

the test statistic is:-

$z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.201-0.18}{\sqrt{\frac{0.18(1-0.18)}{1800}}}=\mathbf{2.32}$

the z critical value for alpha= 0.10, right tailed test be:-

2* = 1.28

decision:-

so, we reject the null hypothesis.

conclusion:-

we conclude that,there is  enough evidence to reject the center’s claim at 0.10 level of significance.

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