Question

In: Statistics and Probability

In a random sample of 100 adult Americans who did not attend college, 37 said they...

In a random sample of 100 adult Americans who did not attend college, 37 said they believe in extraterrestrials. In a random sample of 100 adult Americans who did attend college, 47 said they believe in extraterrestrials. At the 1% level of significance, does this indicate that the proportion of people who did not attend college and believe in extraterrestrials is less than the proportion of people who did attend college and believe in extraterrestrials?

State the null and alternate hypotheses

Identify the type of test (left-tailed, right-tailed, two-tailed) and state the level of significance.

Determine the p-value or the interval containing the p-value and make a decision to either reject/not reject the null hypothesis.

Interpret your decision in the context of the problem.

Solutions

Expert Solution

Here, we have given that,

n1= number of adult Americans who did not attend college=100

x1=number of adult Americans who did not attend college and said they believe in extraterrestrials= 37

\hat p1=Sample proportion of adult Americans who did not attend college and said they believe in extraterrestrials

=

n2= number of adult Americans who did attend college=100

x2=number of adult Americans who did attend college and said they believe in extraterrestrials= 47

\hat p2=Sample proportion of adult Americans who did attend college and said they believe in extraterrestrials

=

Claim: To check whether the population proportion of people who did not attend college and believe in extraterrestrials is less than the population proportion of people who did attend college and believe in extraterrestrials.

The null and alternative hypotheses are as follows:

Versus

Where p1 = The population proportion of people who did not attend college and believe in extraterrestrials

p2= The population proportion of people who did attend college and believe in extraterrestrials.

This is the left one-tailed test.

= level of significance =1%=0.01

Here, we are using the two-sample proportion test to test the hypothesis.

Now, we can find the test statistics

Z-statistics=

                 =

       =

= -1.44

The test statistic is -1.44.

Now, we can find the p-value

P-value = P( Z<  z-statistics) As this is left one tailed test

              = P (Z < -1.44)

              =0.0749 Using standard normal z table see the value corresponding to the z=-1.44

we get the p-value is 0.0749

Decision:

Here, P-value (0.0749) greater than (>) 0.01

Conclusion:

we do not reject the Null hypothesis Ho

Interpretation:

we conclude that there is not sufficient evidence to support the claim the population proportion of people who did not attend college and believe in extraterrestrials is less than the population proportion of people who did attend college and believe in extraterrestrials.


Related Solutions

A random sample of 283 adult Canadians was taken, and 205 of them said that they...
A random sample of 283 adult Canadians was taken, and 205 of them said that they always vote in federal general elections. Find a 92% confidence interval for the proportion of adult Canadians who say that they always vote in federal general elections
1. In a random sample of 200 college graduates, 42 said that they think a college...
1. In a random sample of 200 college graduates, 42 said that they think a college degree is not worth the cost. Test to see if this sample provides significant evidence that the population proportion of college graduates who believe a college degree is not worth the cost is DIFFERENT FROM 25%. Use a 5% significance level. Round all calculations to three decimal places. Verify that the sample size is large enough (2 short calculations) a. Write the null and...
Thirty five percent of adult Americans are regular voters. A random sample of 350 adults in...
Thirty five percent of adult Americans are regular voters. A random sample of 350 adults in a medium size college town were surveyed, and it was found that 155were regular voters. Estimate the true proportion with 95% confidence and comment on your results.Write a conclusion for the confidence interval.Round to 3 decimalplaces.
A Pew Poll conducted in 2017 surveyed a random sample of 200 adult Americans and 75...
A Pew Poll conducted in 2017 surveyed a random sample of 200 adult Americans and 75 of them said that they had seen a ghost. Find a 99% confidence interval for the percentage of all Americans who believe they have seen a ghost. Interpret your interval. Round to three decimal places.
In March 2014, Harris Interactive conducted a poll of a random sample of 2234 adult Americans...
In March 2014, Harris Interactive conducted a poll of a random sample of 2234 adult Americans 18 years of age or older and asked, "which is more annoying to you, tailgaters or slow drivers who stay in the passing lane?" Among those surveyed, 1184 were more annoyed by tailgaters a) Explain why the variable of interest is qualitative with two possible outcomes. What are the two outcomes? b) Verify the requirements for constructing a 90% confidence intervsl for the population...
According to a recent by the CDC, 37% of Americans said that they'd eaten fast food...
According to a recent by the CDC, 37% of Americans said that they'd eaten fast food within the past 24 hrs with 7.8% of Americans being considered High income AND have eaten fast food within the past 24HRS. It's also know that 21.1% of Americans are High Income, 49.9% are Middle Income, and the rest of individuals are Low Income. Finally, 18.0% of Americans are considered Middle Income AND have eaten fast food within the past 24 HRS. A) calculated...
5. In a random sample of 200 adult Americans, 8 had some form of color blindness....
5. In a random sample of 200 adult Americans, 8 had some form of color blindness. a. Construct and interpret a 95% confidence interval using the Wilson-adjusted method. b. Using the information above as a pilot study, how many adult Americans would need to be sampled in order to estimate the population proportion with a standard error of no more than 0.005? c. If no prior estimate of the prevalence of color blindness was available, how many adult Americans would...
Data from a random sample of 2300 adult Americans found that 1300 of those surveyed would...
Data from a random sample of 2300 adult Americans found that 1300 of those surveyed would prefer to live in a hot climate rather than a cold climate. Do the sample data provide convincing evidence that a majority of all adult Americans prefer a hot climate over a cold climate? (a) State the null and alternative hypotheses to be tested. (b) Test the relevant hypotheses using α = .01. (Be sure to check all necessary conditions and clearly state your...
A random sample found that forty percent of 100 Americans were satisfied with the gun control...
A random sample found that forty percent of 100 Americans were satisfied with the gun control laws in 2017. Compute a 99% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2017. Fill in the blanks appropriately. A random sample found that forty percent of 100 Americans were satisfied with the gun control laws in 2017. Compute a 99% confidence interval for the true proportion of Americans who were satisfied with the...
(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was...
(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was fed seeds on 22 days and Parsnip was fed pellets on the other 15 days. The goal is to calculate a 99% confidence interval for the proportion of all days in which Parsnip was fed seeds, and to do so there are two assumptions. The first is that there is a simple random sample, which is satisfied. What is the second assumption, and specific...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT