Answered step by step
Verified Expert Solution
Link Copied!

Question

1 Approved Answer

Hi i need help with these questions. Thank you! You wish to test the following claim (HI) at a significance level of a 0.05. =

Hi i need help with these questions. Thank you!

image text in transcribed

You wish to test the following claim (HI) at a significance level of a 0.05. = 72.2 > 72.2 A news article that you read Stated that 57% of voters prefer the Democratic candidate. You think that th actual percent is larger. 149 of the 244 voters that you surveyed said that they prefer the Democratic candidate. What can be concluded at the 0.05 level of significance? You believe the population is normally distributed, but you do not know the standard deviation. You obtain a. For this study, we use a sample of size n = 23 with mean = 76.1 and a standard deviation of s = 8.4. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic- I What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... O less than (or equal to) a O greater than a This test statistic leads to a decision to. O reject the null O accept the null O fail to reject the null As such, the final conclusion is that... O There is sufficient evidence to warrant rejection of the claim that the population mean is greater than 72.2. O There i not sufficient evidence to warrant rejection of the claim that the population mean is greater than 72.2. O The sample data support the claim that the population mean is greater than 72.2. O There is not sufficient sample evidence to support the claim that the population mean is greater than 72.2. A fitness center is interested in finding a 98% confidence interval for the mean number of days per week that Americans who are members of a fitness club go to their fitness center. Records of 277 members were looked at and their mean number of visits per week was 2.2 and the standard deviation was 2.7. Round answers to 3 decimal places where possible. a. To compute the confidence interval usea ? v distribution. b. The null and alternative hypotheses would be: (please enter a decimal) (Please enter a decimal) c. The test statistic (please show your answer to 2 decimal places.) d. The p-value - (Please show your answer to 4 decimal places.) e. The pvalue is ? v f. Based Oh this, we should the null hvnnthocie g. Thus, the final conclusion is that . C) The data suggest the population proportion is not significantly larger 57% at a = 0.05, so the is not sufficient evidence to conclude that the proportion of voters who prefer the Democrati candidate is larger 57%. C) The data suggest the population proportion is not significantly larger 57% at a = 0.05, so ther is sufficient evidence to conclude that the proportion of voters who prefer the Democratic candidate is equal to 57%. O The data suggest the populaton proportion is significantly larger 57% at a = 0.05, so there is sufficient evidence to conclude that the proportion of voters who prefer the Democratic candidate is larger 57% h. Interpret the level of significance in the context of the study. O There is a 5% chance that the proportion of voters who prefer the Democratic candidate is larger 57%. C) There is a 5% chance that the earth is flat and we never actually sent a man to the moon. O If the population proportion of voters who prefer the Democratic candidate is 57% and if another 244 voters are surveyed then there would be a 5% chance that we would end up falsely concluding that the proportion of voters who prefer the Democratic candidate is larger 57% C) If the proportion of voters who prefer the Democratic candidate is larger 57% and if another 244 voters are surveyed then there would be a 5% chance that we would end up falsely concluding that the proportion of voters who prefer the Democratic candidate is equal to 57%. b. With 98% confidence the population mean number of visits per week is between visits. and c. If many groups of 277 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals willsontain the true population mean number of visits per week and about percent will not contain the true population mean number of visits per week. The average American gets a haircut every 34 days. Is the average smaller for college students? The data includes the results of a survey of 13 college students asking them how many days elapse between haircuts. Assume that the distribution of the population is normal and that we know population standard deviation. Suppose that you are given the test statistic: z = -2.05. Answer the following questions: We should conduct a Select an answer v test. The p-value = (Please show your answer to 4 decimal places.) You are interested in estimating the the mean weight of the local adult population of female white-tailed deer (doe). From past data, you estimate that the standard deviation of all adult female white-tailed deer in this region to be 20 pounds. What sample size would you need to in order to estimate the mean weight of all female white-tailed deer, with a 94% confidence level, to within 4 pounds of the actual weight? Round your critical value to two decimal places. Sample Size:

Step by Step Solution

There are 3 Steps involved in it

Step: 1

blur-text-image

Get Instant Access to Expert-Tailored Solutions

See step-by-step solutions with expert insights and AI powered tools for academic success

Step: 2

blur-text-image

Step: 3

blur-text-image

Ace Your Homework with AI

Get the answers you need in no time with our AI-driven, step-by-step assistance

Get Started

Recommended Textbook for

Algebra

Authors: Marvin L Bittinger

12th Edition

0321922913, 9780321922915

More Books

Students also viewed these Mathematics questions

Question

The fear of making a fool of oneself

Answered: 1 week ago

Question

Annoyance about a statement that has been made by somebody

Answered: 1 week ago