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Answer:
No, the marketing manager was not correct in his claim.
Step-by-step explanation:
We are given that in a survey of 1005 adult Americans, 46.6% indicated that they were somewhat interested or very interested in having web access in their cars.
Suppose that the marketing manager of a car manufacturer claims that the 46.6% is based only on a sample and that 46.6% is close to half, so there is no reason to believe that the proportion of all adult Americans who want car web access is less than 0.50.
<em>Let p = population proportion of all adult Americans who want car web access</em>
SO, Null Hypothesis, : p
50% {means that the proportion of all adult Americans who want car web access is more than or equal to 0.50}
Alternate Hypothesis, : p < 50% {means that the proportion of all adult Americans who want car web access is less than 0.50}
The test statistics that will be used here is<u> One-sample z proportion statistics</u>;
T.S. = ~ N(0,1)
where, = sample proportion of Americans who indicated that they were somewhat interested or very interested in having web access in their cars = 46.6%
n = sample of Americans = 1005
So, <u><em>test statistics</em></u> =
= -2.161
<em>Since in the question we are not given the level of significance so we assume it to be 5%. Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z so we sufficient evidence to reject our null hypothesis as it will fall in the rejection region.</em>
Therefore, we conclude that the proportion of all adult Americans who want car web access is less than 0.50 which means the marketing manager was not correct in his claim.