Question

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. Is the marketing manager correct in his claim? Provide statistical evidence to support your answer. For purposes of this exercise, assume that the sample can be considered as representative of adult Americans. Test the relevant hypotheses using

Answer 1

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.

Let p = population proportion of all adult Americans who want car web access

SO, Null Hypothesis, $H_0$ : p $\geq$ 50%   {means that the proportion of all adult Americans who want car web access is more than or equal to 0.50}

Alternate Hypothesis, $H_a$ : 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 One-sample z proportion statistics;

                T.S.  =  $\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where,  $\hat p$ = 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, test statistics  =  $\frac{0.466-0.50}{\sqrt{\frac{0.466(1-0.466)}{1005} } }$

                               =  -2.161

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.

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.