Question

In: Statistics and Probability

A Rochester newspaper claim states that more than 60% of Rochester residents root for the Buffalo...

A Rochester newspaper claim states that more than 60% of Rochester residents root for the Buffalo Bills football team. A recent survey random survey of 387 Rochester residents showed 254 of them root for the Bills. Is there enough evidence at the 0.05 level of significance to prove the claim made by the newspaper is correct? What is the probability of a Type I Error here?
a. Write the null and alternative hypotheses in Word b. List any necessary requirements in Word .e. What can you conclude about your hypothesis (in words) based on these results?

Solutions

Expert Solution

a) As we are testing here whether more than 60% of Rochester residents root for the Buffalo Bills football team, therefore the null and the alternate hypothesis here are given as:

b) The requirements here is that the population size is large enough which is a valid assumption here. Also that the sample drawn is from the population representative sample.

The sample proportion here is computed as:

p = x/n = 254 / 387 = 0.6563

The test statistic here is computed as:

Now as this is a one tailed test, the p-value here is computed from the standard normal tables as:

p = P(Z > 2.2608) = 0.0119

As the p-value here is 0.0119 < 0.05 which is the level of significance, therefore the test is significant and we can reject the null hypothesis and conclude that we have sufficient evidence that more than 60% of Rochester residents root for the Buffalo Bills football team,


Related Solutions

Government officials claim that no more than 29% of the residents of a state are opposed...
Government officials claim that no more than 29% of the residents of a state are opposed to building a nuclear plant to generate electricity. Local conservation groups claim that the true percentage is much higher. To test the government’s claim, an independent testing group selects a random sample of 81 state residents and finds that 29 of the people are opposed to the nuclear plant. (a) What is the variable? Is it an attribute or measurable? (b) State the hypotheses...
Suppose a newspaper article states that the distribution of auto insurance premiums for residents of California...
Suppose a newspaper article states that the distribution of auto insurance premiums for residents of California is approximately normal with a mean of 1502. The article also states that 10% of California residents pay more than 1940. 1. What is the Z score that corresponds to the top 10% (or the 90th percentile) of the standard normal distribution? Round your answer to 4 decimal places. -.7742 2. What is the mean insurance cost? 3. What is the cutoff for the...
Suppose a newspaper article states that the distribution of auto insurance premiums for residents of California...
Suppose a newspaper article states that the distribution of auto insurance premiums for residents of California is approximately normal with a mean of 1502. The article also states that 25% of California residents pay more than 1875. 1. What is the Z score that corresponds to the top 25% (or the 75thth percentile) of the standard normal distribution? Round your answer to 4 decimal places. 2. What is the mean insurance cost? 3. What is the cutoff for the 75thth...
Suppose a newspaper article states that the distribution of auto insurance premiums for residents of California...
Suppose a newspaper article states that the distribution of auto insurance premiums for residents of California is approximately normal with a mean of 1525. The article also states that 20% of California residents pay more than 1875. 1. What is the Z score that corresponds to the top 20% (or the 80?ℎth percentile) of the standard normal distribution? Round your answer to 4 decimal places. 2. What is the mean insurance cost? 3. What is the cutoff for the 80?ℎth...
Do GA residents (population 1) spend more on cable than FL residents (population 2)? A GA...
Do GA residents (population 1) spend more on cable than FL residents (population 2)? A GA sample of 100 showed an average bill of $120 with a standard deviation of $15. A FL sample of 100 showed an average bill of $110 with a standard deviation of $10. Formulate the hypothesis: A. Ho: μ1–μ2<0 . Ha: μ1–μ2>0 B. Ho: μ1–μ2≤0 . Ha: μ1–μ2>0 C. Ho: μ1–μ2>0 . Ha: μ1–μ2≤ 0 D. Ho: μ1–μ2=0 . Ha: μ1–μ2>0 What is the p-value?...
An article in the The Star Newspaper implies that more than half of all Malaysians would...
An article in the The Star Newspaper implies that more than half of all Malaysians would prefer to being given RM100 rather than a day off for work. This statement is based on a survey conducted by Malaysian Employee Incentives and Remunerations Research Consultant Agency, in which 1,040 of 2,000 respondents indicated that they would rather have the RM100. At 0.05 level of significance, is there evidence based on the survey data that more than half of all Malaysians would...
The claim is that more than 75% of workers are satisfied with their job, and sample...
The claim is that more than 75% of workers are satisfied with their job, and sample statistics include 600 employed adults; with 490 of them saying that they are satisfied with their job. Use a 0.01 confidence interval to test the claim. Claim: Null Hypothesis: Alternative Hypothesis: Test Statistic: P-Value: Interpreting the Test Statistic by using P-Value: Critical Value: Interpreting the Test Statistic by using Critical Value: Conclusion:
A newspaper article claimed that more than 42% of women ages 18-33 are mothers. In a...
A newspaper article claimed that more than 42% of women ages 18-33 are mothers. In a random sample of 150 women ages 18-33, 72 were mothers. Use this random sample to test the newspaper article's claim at the 10% significance level. a: state hypotheses b: verify conditions that hold c: calculate the test statistic d: calculate the p value
There are over 5,000 banks in the United States—more than 10 times more per person than...
There are over 5,000 banks in the United States—more than 10 times more per person than in other industrialized countries. A recent study suggests that the long-run average cost curve for an individual bank is relatively flat. If Congress took steps to consolidate banks (merge some of the banks), thereby reducing the total number to 2,500, what would you expect to happen to average costs within the banking industry?
There are over 5,000 banks in the United States—more than 10 times more per person than...
There are over 5,000 banks in the United States—more than 10 times more per person than in other industrialized countries. A recent study suggests that the long-run average cost curve for an individual bank is relatively flat. If Congress took steps to consolidate banks, thereby reducing the total number to 2,500, what would you expect to happen to costs within the banking industry? Please provide a detailed explanation in terms of what long-run average cost curve for the individual bank...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT