Sample surveys show that fewer people enjoy shopping in stores than the past. A recent survey asked a random sample of 2500 U.S. adult if they agreed or disagreed with the statement, "I prefer to shop online rather than in a store." In this survey, 1575 agreed. A manager of a local clothing store claims that 60% of all U.S. adults would say "Agree if asked the same question."

Questions:

1) If the manager's claim is true, what is the probability that the proportion in a random sample of 2500 U.S. adults is at least as far above as 0.60 as the results of this survey?

2) Based on your answer in question 1, do you have reaon to doubt the managers claim?

2) For this case since the probability is very high we can conclude that the manager is correct since is very improbable that the true proportion would be lower than 0.6.

Explanation:

For this case we have a randomsample of 2500 people, this value represent n.

We know that in the sample selected the number of people who say that prefer shop online rather than in a store are 1575.

Part 1

For this case we can calculate the sample proportion with this formula:

[tex] hat p =\frac{X}{n}= \frac{1575}{2500}=0.63[/tex]

Now we can calculate the probability since we assume that the distribution for p is given by:

[tex] p \sim N (\hat p, \sqrt{\frac{\hat p (1-\hat p)}{n}})[/tex]

If we find the mean and the deviation we got:

[tex] \mu_p = 0.63[/tex]

[tex] \sigma_p =\sqrt{\frac{0.63*(1-0.63)}{2500}}= 0.00966[/tex]

And we want to calculate the following probability:

[tex] P(p>0.6)[/tex]

And we can use the z score formula given by:

[tex] z= \frac{p -\mu_p}{\sigma_p}[/tex]

And if we replace we got:

[tex] P(p>0.6) = 1-P(p<0.6) = 1-P(Z<\frac{0.6-0.63}{0.00966}) = 1-P(Z<-3.106)=1-0.000948=0.999[/tex]

Part 2

For this case since the probability is very high we can conclude that the manager is correct since is very improbable that the true proportion would be lower than 0.6.