An article predicts that "spit," spam that is delivered via internet phone lines and cell phones, will be a growing problem as more people turn to web-based phone services. In a poll of 5500 cell phone users, 19% indicated that they had received commercial messages and ads on their cell phones. Is there sufficient evidence that the proportion of cell phone users who have received commercial messages or ads in 2004 was greater than the proportion of 0.13 reported for the previous year? (Use α = 0.05. Round your test statistic to two decimal places and your P-value to four decimal places.)z =
P =There is evidence to suggest that the proportion cell phone users who have received commercial messages or ads in 2004 is greater than the proportion of 0.13 reported for the previous year.

We conclude that the proportion of cell phone users who have received commercial messages or ads in 2004 was greater than the proportion of 0.13 reported for the previous year.

Step-by-step explanation:

We are given that in a poll of 5500 cell phone users, 19% indicated that they had received commercial messages and ads on their cell phones.

We have to test the claim that the proportion of cell phone users who have received commercial messages or ads in 2004 was greater than the proportion of 0.13 reported for the previous year.

Let p = proportion of cell phone users who have received commercial messages or ads in 2004.

So, Null Hypothesis, : p 0.13      {means that the proportion of cell phone users who have received commercial messages or ads in 2004 was smaller than or equal to the proportion of 0.13 reported for the previous year}

Alternate Hypothesis, : p > 0.13      {means that the proportion of cell phone users who have received commercial messages or ads in 2004 was greater than the proportion of 0.13 reported for the previous year}

The test statistics that would be used here One-sample z proportion statistics;

                     T.S. =    ~ N(0,1)

where, = sample proportion of cell phone users who have received commercial messages or ads in 2004 = 19%

           n = sample of cell phone users = 5500

So, test statistics  =  

                               =  11.34

The value of z test statistics is 11.34.

Also, P-value of the test statistics is given by;

         P-value = P(Z > 11.34) = 1 - P(Z 11.34)

                                             = 1 - 0.9999 = 0.0001

Now, at 0.05 significance level the z table gives critical value of 1.645 for right-tailed test.

Since our test statistics is way more than the critical value of z as 11.34 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the proportion of cell phone users who have received commercial messages or ads in 2004 was greater than the proportion of 0.13 reported for the previous year.

step-by-step explanation:

30% of $420 is 0.30($420) = $126 and $420 + $126 = $546.

or, 130% of $420 is 1.3($420) = $546.

notice that this is not the $600 original list price with which we started.

why doesn't this work?

the method does not work because 30% of $420 is not the same amount as the 30% of $600. it may seem obvious when stated this way, but it is a very common error made on problems of this type. we cannot just use the same percent and get back to the original amount because we are taking that percent of a different value.

to work problems of this type correctly, it is generally best to write an equation for the situation and then solve.

example: since (100% – 30%)(original price) = (sale price)

70% of the   (original price) is $420.

0.7x   = 420

equation image indicator

x = 600

the original cost was $600.

we may also solve this problem using the two-step method. working the problem in two steps requires us to use more algebra skills. we have to know that x = 1x (identity property of multiplicationtext annotation indicator) and that 1x – 0.3x = 0.7x (subtracting like terms). then we solve the problem as follows:

(original price) – 30%(original price) = sale price

x – 0.3x = 420

0.7x = $420

equation image indicator

x = 600

the original cost was $600.  

note that with both of the above methods we do obtain the correct original amount.

example: after a 5% pay raise, hermione is earning $22,680 per year. what was she earning before the pay raise?

105% of (original salary) = current salary

1.05x = 22,680

equation image indicator

x = 21,600

hermione earned $21,600 per year before the pay raise.

this is an example so maybe you will get how to at least do it!

(-1,-0.5)

step-by-step explanation: