s In a USA Values a USA Today/Gallup poll, 768 of 1024 In domly selected adult a

Question

s In a USA Values a USA Today/Gallup poll, 768 of 1024 In domly selected adult aged 18 older stated that candidate's positions on the issue of value extremely a important in determining their vote for president or very ain point estimate for the proportion of adult Americans aged 18 or older for which the issue of family values is extremely or very important in determining their vote for president. ih) verify that the requirements for constructing a confidence interval for p are satisfied. (c) Construct a 99% confidence interval for the proportion of adult Americans aged 18 or older for which the issue of N family values is extremely or very important in determining their vote for president. (d) Is it possible that the proportion of adult Americans aged 18 or older for which the issue of family values is extremely or very important in determining their vote for president is below 70% Is this likely?

Explanation / Answer

1) Here, from the sample proportion , the point estimate comes to p= 768/1024 = 0.75 or 75% of the population.

This is the required point estimate of the population.

2) The statistc which the proportion of people who conform to a the above view: can be said to be normally distributed when the sample size is large ie if np and n(1-p) are both greater than 10 , then we can safely assume that the proportion is normally distributed. Here it is 1024*0.75=768 and 1024*0.25= 256 both of which is >>10.

Therefore we can apply the central limit theorem for proportions.

3)

The 99% confidence interval for the sample proportion is given by:

p+ z*sqrt(p(1-p)/n)

0.75+2.575*sqrt((0.75*0.25)/1024)= 0.78

p- z*sqrt(p(1-p)/n)= 0.71

Therefore the 99% confidence interval for true proportion lies between 71% and 78%

4) Here our null hypothesis is p>= 0.70

ha: p<0.70

Now we have the liberty to decide the decison rule to accept or reject the hypothesis based on the type 1 error rate.(alpha)

If we assume alpha=0.025

we get z statistic value for our left tailed test as -1.96. we reject null hypothesis if our test statistic falls below -1.96

Calculating test statistic:

p-po/(sqrt(po*(1-po)/n))

0.75-0.70/(sqrt(0.7*0.3)/1024)

We dont reject the null hypothesis in this case.

Also from the previous question we saw that there is 99% chance that it falls in the interval mentioned(71 and 78).

It is very unlikely to fall below 70%

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