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An e-commerce research company claims that 60% or more graduate students have bought merchandise on-line. A consumer group is suspicious of the claim and thinks that the proportion is lower than 60%. A random sample of 80 graduate students show that only 22 students have ever done so. Is there enough evidence to show that the true porportion is lower than 60%? Use a significance of .05. Show your null and alternative hypotheses, your p-value and state your conclusion in context of the problem.

This question involves hypothesis testing using the z distribution, the formula z = (p^ - p)/(p*(1-p)/n)

where p^ is the observed sample proportion, p is the hypothesised proportion and n is the sample size used. 

p^ = 22/80 = 0.275 and p is the hypothesised proportion (0.60), 


H0: p = 0.60

Ha: p < 0 xss=removed>

You can lookup the p value from the standard normal distribution table or use the Excel's NORMS.S.DST() function,  =NORM.S.DIST(-5.93,TRUE) , The p value is approximately 0, alternatively, you can compare the Test statistic obtained and the z score value associated with an alpha of 0.05, (The left tailed alpha), =NORM.S.INV(0.05) = -1.645

Reject the null hypothesis because the Test statistic is less than -1.645, There is not enough evidence to support the company's claim, 





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