The number y of hits a new website receives each month can be modeled by y = 4020ekt, where t represents the number of months th
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Answer:
No, there is not enough evidence at the 0.02 level to support the executive's claim.
Step-by-step explanation:
We are given that a publisher reports that 34% of their readers own a particular make of car. A random sample of 220 found that 30% of the readers owned a particular make of car.
And, a marketing executive wants to test the claim that the percentage is actually different from the reported percentage, i.e;
Null Hypothesis, : p = 0.34 {means that the percentage of readers who own a particular make of car is same as reported 34%}
Alternate Hypothesis, : p
0.34 {means that the percentage of readers who own a particular make of car is different from the reported 34%}
The test statistics we will use here is;
T.S. = ~ N(0,1)
where, p = actual % of readers who own a particular make of car = 0.34
= percentage of readers who own a particular make of car in a
sample of 220 = 0.30
n = sample size = 220
So, Test statistics =
= -1.30
Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.
Therefore, we conclude that the actual percentage of readers who own a particular make of car is same as reported percentage and the executive's claim that it is different is not supported.