Question

2) At a trade show, you interview a random sample of 50 attendees. The results of the survey show that 48% of the attendees said they were more likely to visit an exhibit when there is a giveaway. At α=0.05 test the claim that at least 52% of the attendees at trade shows are more likely to visit an exhibit when there is a giveaway.

e. Please state both the null and alternative hypotheses.

f. Please write a decision rule that states when the investigator should reject the null hypothesis.

g. Please show the R code and results needed to generate the appropriate statistic.

h. State your decision as to whether the investigator should reject the null hypothesis.

8.0.3

Answer 1

e. The null and alternative hypothesis is ,

The test is one-taile test.

f. The critical value is ,

Decision rule : If Z-stat<-1.64 , then reject Ho , otherwise do not reject.

g. The R-code is ,

> prop.test(x=50*0.48,n=50,p=0.52,conf.level=0.95,,correct=FALSE)

1-sample proportions test without continuity correction

data: 50 * 0.48 out of 50, null probability 0.52
X-squared = 0.32051, df = 1, p-value = 0.2856
alternative hypothesis: true p is less than 0.52
95 percent confidence interval:
0.0000000 0.5942248
sample estimates:
p
0.48

h) The test statistic is ,

Decision : Here , Z-stat=-0.566>-1.96

Therefore , Do not reject Ho.

Conclusion : There is sufficient evidence to support the claim that at least 52% of the attendees at trade shows are more likely to visit an exhibit when there is a giveaway.